§1. Sequences
Consider the function f with domain the set of nonnegative integers, IN = {0, 1, 2, . . .},
which is defined as follows:
i) f (0) = 6,
ii) f (n) = 2f (n − 1) − 5 for each positive integer n.
This is an example of a recursive definition of a function, meaning that first a finite set
of the values are given, in this case just the value at 0, and then the remaining values are
defined in terms of one or more previous values.
We can use this definition to compute as many values of f as we wish by starting with
the value f (0) = 6 given in part (i), and then using part (ii) repeatedly. At the beginning
we obtain
f (1) = 2f (0) − 5 = 2 · 6 − 5 = 7,
f (2) = 2f (1) − 5 = 2 · 7 − 5 = 9,
f (3) = 2f (2) − 5 = 2 · 9 − 5 = 13,
f (4) = 2f (3) − 5 = 2 · 13 − 5 = 21.
Continuing in this way would generate a list of values of f which begins with
6, 7, 9, 13, 21, 37, 69, 133, 261, 517, . . . .
The entire infinite list of values in the proper order would give an alternate definition of
the function f .
Such an ordered list is an example of an infinite sequence. The locations of the
individual terms of the sequence are usually indicated by using the members of IN as
subscripts, as in
a0 , a1 , a2 , a3 , . . . , an , . . . .
The entire sequence is then often denoted by
(an ) or (an )n≥0
or (an )n=∞
n=0 .
By an infinite sequence we always mean just such an ordered list.
Thus each func-
tion with domain IN determines an infinite sequence of its values and each sequence
1
§1. Sequences
a0 , a1 , a2 , . . . , an , . . . defines a function f on IN, where f (n) = an for integers n ≥ 0.
(In some cases we may begin a sequence with a1 instead of a0 , corresponding to a function
on the positive integers, or even start with some am , m > 1, corresponding to a function
on {m, m + 1, m + 2, . . .}.)
Example 1.1. In dealing with sequences we will usually want some way to compute the
nth term, an , (actually the (n + 1)st if we start with a0 , but this is what people call an )
from some general formula. For example, consider the sequence (bn ) which begins with
b0 = 1, b1 = 2, b2 = 4, b3 = 8, and which is defined recursively by
i) b0 = 1
ii) bn = 2bn−1 for n ≥ 1.
Here it is easy to see that the nth term is given by bn = 2n for each integer n ≥ 0. To prove
this it is enough to check that this formula is correct for 0 since b0 = 1 = 20 , and that if it
is correct for any integer n − 1 so that bn−1 = 2n−1 , then it is also true for the next integer
n since bn = 2bn−1 = 2 · 2n−1 = 2n . (This is the essence of a proof by induction.)
If you write out a few more terms of the sequence (bn ) with bn = 2n , you may note a
similarity to the sequence of values of the first example:
(bn ) 1, 2, 4, 8, 16, 32, 64, 128, 256, 512,. . .
(an ) 6, 7, 9, 13, 21, 37, 69, 133, 261, 517,. . .
Both of the sequences are increasing and, at least so far, seem to be doing so at about
the same rate. Looking more closely you may see that each term in the second sequence is
exactly 5 greater than the corresponding term of the first. For these terms then, we have
an = 2n + 5.
Does this remain so for all n ≥ 0? We can prove that it does by arguing “inductively.”
First we use the fact that it is correct at the start: a0 = 20 +5 = 6. The recursive definition
of the function f gives an = 2an−1 − 5. This shows that if the formula is correct for an
integer n − 1 with an−1 = 2n−1 + 5, then it remains correct for an since
an = 2an−1 − 5 = 2(2n−1 + 5) − 5 = 2n + 5.
2
§1. Sequences
When we have a sequence which is defined recursively (or when we know the first
several terms and can assume that some pattern is continued for the whole sequence) it
may not be possible to find a formula for the nth term just by “noticing” a relationship to
some other sequence or by guessing. If we have come up with the correct formula, then in
most cases we can verify that it does hold for the entire sequence by an inductive argument
such as those in the previous examples. But first we need to find such a formula. Methods
for doing so will be one of our major topics.
As with the two examples so far, most of the sequences we will be dealing with will be
increasing. In a typical case the nth term, an , might give the number operations required
when a particular algorithm is used to solve a problem with input which, by some measure,
has size n. We would expect the result to be an increasing sequence in such a case. If we
considered two different algorithms for solving the same problem, we would have two such
sequences, and would be interested in comparing the rates at which they increase. A slower
rate of increase would correspond to a more efficient algorithm. To compare increasing
sequences (an ) and (bn ) we will consider the ratios of corresponding terms. That is, we
will investigate the sequence (cn ), where
cn =
an
.
bn
Comparing the two sequences above in this way would involve considering the sequence
with
an
2n + 5
cn =
=
.
bn
2n
This sequence of ratios is not increasing. In fact, since
2n + 5
5
cn =
= 1+ n,
n
2
2
it is easy to see that as n increases the term 5/2n gets smaller, and cn decreases towards 1.
By going far enough in the sequence we could insure that for all remaining terms 5/2n is as
close to 0 as we might wish, and therefore that cn would remain this close to 1 from then
on. This is the key idea behind saying that in this case the sequence (cn ) has a finite limit
which is 1. To go very far with this idea we need some formal definitions. Before stating
them it may be useful to consider some more examples.
3
§1. Sequences
There are no candidates for finite limits for the increasing sequences given by an =
2n + 5 or bn = 2n , but an increasing sequence might have a finite limit as in the case of
the next example.
Example 1.2. an = 1 − 1/n (n ≥ 1) : 0, 1/2, 2/3, 3/4, 4/5, . . . , 1 − 1/n, . . . .
Here an differs from 1 by 1/n which decreases toward 0 as n increases. We want a definition
which implies that this sequence has 1 as a limit. Here, although this is an increasing
sequence – each term is larger than the one before it – the terms do not get arbitrarily
large. We have an < 1 for every n ≥ 0. We will see that this combination, increasing but
“bounded” from above, always insures a limit.
Example 1.3. The sequence given recursively by a0 = 1 and an = 1/3an−1 for all n ≥ 1.
This sequence has an = ( 1/3)n as its nth term for all n ≥ 0. (Verify this by an inductive
argument.) So this is a decreasing sequence, whose terms are bounded from below since
an ≥ 0 for all n ≥ 0. If any positive distance were specified, by going far enough in the
sequence we could insure that all remaining terms would be within that distance of 0. This
is the key idea behind saying that this sequence converges to the limit 0.
Example 1.4. Finally, consider the sequence for which an = (−1)n for n ≥ 0. This gives
1, −1, 1, −1, 1, −1, . . . , (−1)n , . . . .
It is true that no matter how far we go in this sequence we will always continue to find
terms close to 1 – in fact there are infinitely many more that are equal to 1 – but we will also
always have more terms which differ from 1 by an amount, 2, which is not decreasing. We
will only say that 1 is the limit of a sequence if for every positive distance, no matter how
small, we can go far enough so that all of the remaining terms are within that distance
of 1. We cannot accomplish that here for any distance less than 2. The value −1 will
not serve as a limit for similar reasons, and there is no other value which the terms are
approaching, so this sequence has no limit. When there is no finite limit, as in this case
or for the examples in which an continues to grow without bound such as an = 2n , we say
that the sequence diverges or that it is a divergent sequence.
4
§1. Sequences
Now for a formal statement of these ideas. When speaking of a number L as the
limit of the sequence (an ) we need to consider the distances of terms of the sequence from
L. The difference of two real numbers tells us how far apart they are. To avoid separate
cases which depend on which of the numbers is larger, we use the absolute value of that
difference. The distance between some an and L is the absolute value of the difference,
|an − L|. When this distance is less than some number we can express this fact as an
inequality:
|an − L| < .
To indicate that such an inequality holds for all terms of the sequence from some point
on we can specify the subscripts of the terms for which this is true. The terms from some
particular term aN on are simply those with subscripts n which satisfy n ≥ N . This allows
us to formalize the idea of the limit of a sequence as follows:
Definition 1.1. We say that the sequence (an ) converges to the limit L if, given any
positive number , there exists an integer N such that for each n ≥ N we have |an − L| < .
In this case we write
lim an = L.
n→∞
Example 1.5. Consider the sequence (an ) with an = 1/n for n ≥ 1. It is not hard to show
that this sequence converges to the limit 0.
We cannot compute how large N must be, that is, how far in the sequence we must go
to insure that |an − 0| < by treating each possible positive number separately. There
are infinitely many such choices for . But we can get an idea about how to construct
an argument which would be valid for every positive at once by considering an example
first. Consider the case of = 0.1. The object is to show that there is an integer N such
that if n ≥ N , then
|an − 0| = | 1/n − 0| = | 1/n| = 1/n < 1/10.
Obviously, if n ≥ 11, then 1/n < 1/10, so we could choose N = 11. (We could also choose
N = 12, or 13, or any larger integer – we only need to know that there is a “far enough,”
not the earliest such.)
5
§1. Sequences
For = 0.05 we would need to obtain 1/n < 1/20 for large enough n. Here we could
choose N = 21. In general we would have 1/n < provided that n > 1/, so we can use any
integer N which is sure to be larger than 1/, like N = 21 when 1/ = 20, or N = 34 when
= 0.03 and 1/ =
100/3
= 33.3̄. For any positive number we could add 1 to 1/, and then,
since this may not be an integer, take N to be the integer we obtain by rounding up, that
is, N = d 1/ + 1e. (dxe denotes the least integer greater than or equal to x.)
The general argument would then be as follows: For any positive number we could
choose N = d 1/ + 1e. If n ≥ N , then n ≥ d 1/ + 1e ≥ 1/ + 1. So n − 1 ≥ 1/ and 1/n−1 ≤ .
Then we have
|an − 0| = | 1/n| = 1/n < 1/n−1 ≤ .
Fortunately, we will not often need to go through the details of such a proof. The formal
definition will allow us to establish some general facts about limits of sequences, and these
will be used to handle most situations. Occasionally we will encounter something new and
may be forced back to the definition to sort things out. First, however, we need to compile
a list of some of the basic properties of limits. This is the subject of the next section.
Exercises
1.1. Use the definition to show that a sequence cannot have two different limits.
1.2. The sequence (an ) with an = 1/2n , n ≥ 0 has limit 0. Find an N such that |an −0| < 0.1
for all n ≥ N .
1.3. i) Suppose the sequence (an ) has a limit. Prove that the sequence (|an |) also has a
limit, or give a counterexample.
ii) Suppose the sequence (|an |) has a limit. Prove that the sequence (an ) also has a
limit, or give a counterexample.
1.4. Which of the following sequences converge? (No proofs required, but consider the
definition when trying to decide.)
i) an = 1/n2 , n ≥ 1
ii) an = n/n−1, n ≥ 2
2
iii) an = n /n+1, n ≥ 0
iv) a0 = 1 and an = 1/2an−1 + 1 for n ≥ 1
2
v) an = n −1/2n2 +7, n ≥ 0
vi) an = (−2)n , n ≥ 0
6
§1. Sequences
1.5. Suppose the sequence (an ) has a limit. Prove that the sequence (bn ), where bn = an +1
for each n ≥ 0 has a limit, or give a counterexample.
1.6. Suppose the sequence (an ) has a limit. Prove that the sequence (bn ) = 2an for each
n ≥ 0 has a limit, or give a counterexample.
7
§2. Properties of Limits of Sequences
§2. Properties of Limits of Sequences
As one more illustration of the formal definition of the limit of a sequence, consider
the case of a constant sequence. That is, suppose c is a fixed constant and that a n = c for
each n ≥ 0. If our definition is at all reasonable, it should imply limn→∞ an = c. To check
that (an ) does have c as its limit we would need to show that for each positive number there is an integer N such that |an − c| < for each n ≥ N . In this case we have
|an − c| = |c − c| = |0| = 0 < for each term of the sequence and every positive number . In most cases when a limit
does exist our choice of N would depend on the size of . Here we could take N to be zero
in all cases.
Now consider the sequence for which an = 10 + 1/n for n ≥ 1. By noting that
|an − 10| = |(10 + 1/n) − 10| = | 1/n| = 1/n
we could show that limn→∞ (10 + 1/n) = 10 using much of the same reasoning that verified
limn→∞ ( 1/n) = 0. Rather than repeat those details, we could think of this sequence as
having been formed by adding corresponding terms of the constant sequence bn = 10
and the sequence cn = 1/n for n ≥ 1. We already know that limn→∞ bn = 10 and
that limn→∞ 1/n = 0. All we need in this case is that limn→∞ (bn + cn ) = limn→∞ bn +
limn→∞ cn = 10 + 0 = 10.
That we can in fact add limits this way (when the sequences involved do have limits)
is one of the properties which are collected together in the next statement.
Theorem 2.1. Suppose (an ) and (bn ) are convergent sequences with limn→∞ an = a and
limn→∞ bn = b. Then
i) limn→∞ (an + bn ) = a + b and limn→∞ (an − bn ) = a − b,
ii) for any constant c, limn→∞ (can ) = ca,
iii) limn→∞ an bn = ab,
iv) if b 6= 0, then limn→∞ (an /bn ) = a/b.
8
§2. Properties of Limits of Sequences
This statement should be understood as meaning that the new sequences (can ), (an +
bn ), etc. will be convergent as well as telling what their limits will be.
To see how these parts follow from the definition, consider (ii). We assume that
limn→∞ an = a. We must show that for each positive there is an N such that |can −ca| < for all n ≥ N . Using facts about absolute values gives
|can − ca| = |c(an − a)| = |c||an − a|.
If c = 0, then (can ) is a constant sequence with limit 0 = 0 · a. So we may assume c 6= 0.
Since limn→∞ an = a, we can insure that |an − a| is as small as we wish by going far
enough in the sequence (“there is an N . . .”). Now we need to make |can − ca| small as
well. For any > 0 we will have |can − ca| = |c||an − a| < if |an − a| < /|c|. But /|c| is
also some positive number. For each such there is an integer N which will do the job. So
there exists an integer N such that |an − n| < /|c| and |can − ca| < for all n ≥ N .
The first statement of (i) is not much harder to check. Again it requires a fact about
absolute values, this time the “triangle inequality,” which gives us
|(an + bn ) − (a + b)| = |(an − a) + (bn − b)| ≤ |an − a| + |bn − b|.
The basic idea is that since limn→∞ an = a and limn→∞ bn = b, if we go far enough we
can arrange to have |an − a| small and |bn − b| small as well. Then the inequality can
be used to insure that the left-hand-side is also small. More precisely, we would have
|(an + bn ) − (a + b)| < if we can arrange to have both |an − a| < /2 and |bn − b| < /2
beyond some point in the sequence. We can therefore reason as follows. For each positive
number the number /2 is positive as well. Since limn→∞ an = a, there is an integer N1
such that |an − n| < /2 for each n ≥ N1 , and since limn→∞ bn = b there is an integer N2
such |bn − b| < /2 for all n ≥ N2 . If we insure n is as large as N1 and also as large as
N2 , then both inequalities will hold at once. We can do this by considering those n with
n ≥ N , where N = max(N1 , N2 ) (or N1 + N2 , or some other choice which is certain to be
as large as N1 and as large as N2 ). Now for each n ≥ N we have
|(an + bn ) − (a + b)| ≤ |an · a| + |bn − b| < /2 + /2 = .
9
§2. Properties of Limits of Sequences
The proofs of the other parts of the theorem are somewhat more complicated, but are the
same in spirit.
Note that no claim is made in part (iv) if limn→∞ bn = 0. This is not just because
there is no finite value for a/b. If for example, (an ) and (bn ) are sequences of positive
numbers with (an ) converging to a positive limit while limn→∞ bn = 0, then the terms of
( an/bn ) will continue to grow without bound. Think of an = 1 and bn = 1/n for n ≥ 1. If,
on the other hand, limn→∞ an = 0 and limn→∞ bn = 0, there might be a finite limit (e.g.,
an = bn = 1/n) or no limit (e.g., an = 1/n, bn = 1/n2 ).
2n3 +5
n4 −1
Example 2.1. Consider the sequence
. It might appear that Theorem 2.1 would
be of no use here since neither of the sequences (2n3 + 5) or (n4 − 1) has a finite limit,
and the theorem deals with combinations of convergent sequences. But notice that for any
n ≥ 1 we have
1/n4 2n3 + 5
2/n + 5/n4
2n3 + 5
=
=
.
·
1/n4
n4 − 1
n4 − 1
1 − 1/n4
The new numerator and denominator represent convergent sequences and (by part (ii) of
the Theorem) we have
lim ( 2/n + 5/n4 ) = 0 + 0 = 0,
n→∞
and
lim (1 − 1/n4 ) = 1 − 0 = 1,
n→∞
which by part (iv) gives
lim
n→∞
2n3 + 5
n4 − 1
= lim
n→∞
2/n
+ 5/n4
1 − 1/n4
= 0/1 = 0.
Similar reasoning yields
2n2 + 5
3n2 − 1
2 + 5/n2
lim
= lim
= 2/3.
n→∞
n→∞ 3 − 1/n2
n3 −1
On the other hand, the terms of the sequence 3n
can be rewritten for n ≥ 1 as
2 +4
n − 1/n2
3 + 4/n2
or as
1 − 1/n3
,
3/n + 4/n3
from which it is not hard to see that this sequence will have no finite limit. (For large n
its terms are closely approximated by n/3 which grows without bound.)
10
§2. Properties of Limits of Sequences
We will need to consider such ratios when comparing the “rates of growth” of sequences
which individually have no finite limits.
Part (ii) of Theorem 2.1 can be viewed as a special case of a more general principle.
This is the part which states that if each term of a sequence with limit a is multiplied by
the constant c, the result is a convergent sequence whose limit is obtained by multiplying
a by c. That is,
lim (can ) = c
n→∞
lim an = ca.
n→∞
Multiplication by a constant is a continuous operation. By this we mean that the function
f (x) = cx is continuous for all x. Similarly, the function f (x) = x2 is continuous for all x,
and we could therefore show that if limn→∞ an = a, then limn→∞ (a2n ) = a2 . The general
principle we referred to is the following statement.
Theorem 2.2. Suppose (an ) is a convergent sequence with limn→∞ an = a and that f is
a function which is continuous at the points of some open interval which contains a. Then
the sequence (f (an )) is convergent and we have
lim (f (an )) = f
n→∞
lim an = f (a).
n→∞
The idea here is that by the continuity of f we can insure that f (an ) is as close
as we wish to f (a) for all an sufficiently close to a. And since (an ) converges to a, we
will have the an close to a if n is large enough. More precisely f is continuous at a
point a in its domain provided that the limit of f as x approaches a exists and is f (a).
That is limx→a f (x) = f (a). This in turn means that for each positive number there
is a positive number δ such that |f (x) − f (a)| < whenever |x − a| < δ. To show that
limn→∞ (f (an )) = f (a), we need to establish for an arbitrary > 0 that |f (an ) − f (a)| < when n is large enough. The continuity of f insures that there is a δ > 0 such that this
will be the case if |an − a| < δ. And since (an ) converges to a, there is an integer N such
that |an − a| < δ for all n ≥ N .
Example 2.2.
i) For every positive integer q the function f (x) = xq is continuous on the whole real
line. Therefore if limn→∞ an = a, then limn→∞ (aqn ) = aq .
11
ii) The function f (x) =
§2. Properties of Limits of Sequences
√
x is continuous at all points of its domain, so if (an ) is a
√
sequence of nonnegative numbers with limn→∞ an = a, we also have limn→∞ an =
√
a. (Note that (an ) cannot have a negative limit if an ≥ 0 for all n.)
iii) The absolute value function f (x) = |x| is continuous on the whole real line. (We do
have limx→0 f (x) = 0 = f (0), although f 0 does not exist at x = 0.) Therefore, if
limn→∞ an = a, then limn→∞ (|an |) = |a|.
iv) The trigonometric functions f (x) = sin x and f (x) = cos x are continuous for all x.
Thus if limn→∞ an = a, we have limn→∞ (sin(an )) = sin a and limn→∞ (cos(an )) =
cos a. For example, limn→∞ (cos( 1/n)) = cos |0| = 1.
We noted at the beginning that each sequence can be viewed as a list of the values of
some function on the nonnegative integers or on the integers starting with some positive
value. For example, the sequence an = 1/n consists of the values of f (x) = 1/x at the
members of {1, 2, 3, . . .}. It was not hard to check that limn→∞ ( 1/n) = 0, but this could
also be viewed as a consequence of the fact that the limit of f (x) = 1/x as x → ∞ exists
and is zero. A function f has a finite limit L as x → ∞, limx→∞ f (x) = L, provided
that for each > 0 there is a number c such that |f (x) − L| whenever x ≥ c. If (an )
is the sequence of values of f at the integers beyond some point, so that an = f (n),
this means that |f (n) − L| = |an − L| < whenever n ≥ c. Thus for any integer N
which is as large as c or larger we have |an − L| < for all n ≥ N , which shows that
limn→∞ an = limx→∞ f (x) = L. This is a proof of the following statement.
Theorem 2.3. If for the function f we have limx→∞ f (x) = L and an = f (n) for each
integer n beyond some point, then limn→∞ an = L.
Example 2.3. Most proofs of the fact that the derivative of sin x is cos x make use of the
limit limh→0 sin(h)/h = 1. If we set x = 1/h, then “h → 0” corresponds to “x → ∞” and we
also have limx→∞ sin( 1/x)/ 1/x = limx→∞ x sin( 1/x) = 1. It follows that limn→∞ n sin( 1/n) =
1.
Note that when an = f (n) the sequence (an ) may have a limit even when f (x),
because of its behavior between integer points, has no finite limit as x → ∞. For example,
12
§2. Properties of Limits of Sequences
f (x) = sin(2πx) has no limit as x → ∞, but an = sin(2πn) is 0 for all n ≥ 0, and so has
limit 0.
Exercises
2.1. Determine whether the sequence has a limit, and if so, find it.
i) an = 2n−4
n+1 , n ≥ 0
ii) an = 1 + 3/n2 , n ≥ 1
iii) an = (− 1/2)n , n ≥ 0
iv) an = 1/n − 1/n+1, n ≥ 1
,n≥0
v) an = n24n
+1
vi) an = (1 − ( 1/2)n ) 2n−4
n+1
2.2. Given an example of two divergent sequences (an ) and (bn ) such that
i) (an + bn ) converges
ii) (an + bn ) diverges.
2.3. Show that if (an ) and (bn ) are convergent sequences with an ≥ bn for each n ≥ 0,
then limn→∞ an ≥ limn→∞ bn .
2.4. Show that if limn→∞ an = a 6= 0 and limn→∞ bn = 0, then the terms of (an /bn ) grow
without bound. That is, for each number B and integer N there is an n ≥ N for
which an /bn ≥ B.
2.5. Find limn→∞ an , where
i) an = e−1/n
1/2
2
.
ii) an = n3n−5n
2 +1
13
§3. Bounded Sequences
§3. Bounded Sequences
A sequence (an ) is said to be increasing if an+1 ≥ an for each n, and decreasing
if an+1 ≤ an for each n. Note that this terminology, which is quite standard, has the
somewhat awkward result that a constant sequence, for example, is called “increasing.”
We only required that an+1 be at least as large as an . Similarly, a constant sequence
qualifies as “decreasing.” If the strict inequality an+1 > an holds for all n, then (an ) is
strictly increasing, and it is strictly decreasing if an+1 < an for each n.
We began the first section by considering the sequence an = 2n + 5 which is clearly
increasing. It is equally clear that this sequence could not have a finite limit. For any
positive number M we will have an > M for n large enough. On the other hand, the
sequence an = 1− 1/n is also increasing, and in this case there is a limit, namely, limn→∞ (1−
1/n)
= 1. The crucial difference is that while for an = 1 − 1/n each term is (strictly) larger
than its predecessor, none of the terms is larger than 1.
We say that a sequence (an ) is bounded above if there is a number B such that an ≤ B
for each term an . It is bounded below if there is a number A such that an ≥ A for each an .
The question of whether an increasing sequence converges is entirely a matter of whether or
not it is bounded above. For a decreasing sequence it is a matter of whether it is bounded
below. This is the meaning of the next statement.
Theorem 3.1. An increasing sequence which is bounded above converges. If an ≤ B for
each term of the convergent sequence (an ), then limn→∞ an ≤ B.
A decreasing sequence (an ) which is bounded below converges. If an ≥ A for each
term of the convergent sequence (an ), then limn→∞ an ≥ A.
Suppose (an ) is a sequence with an ≤ B for each n. Then we call B an upper bound
for (an ). Similarly, A is a lower bound for (an ) if an ≥ A for each n. Theorem 3.1 tells us
that an increasing sequence either has some upper bound, and hence a limit, or else just
keeps growing beyond any possible bound. A decreasing sequence has a lower bound and
must converge, or takes on larger and larger negative values.
If (an ) has one upper bound, it has infinitely many. If an ≤ B for all n, then we also
have an ≤ 1.01B, an ≤ 3/2B, an ≤ B + 1, etc. for all n. The truth of Theorem 3.1 rests
14
§3. Bounded Sequences
on the fact that if (an ) has an upper bound, then it has a least upper bound – one which
is less than or equal to all others. This in turn is the result of a fairly deep property of
the real numbers known as completeness. (We must be careful here. Not every set of real
numbers, even when bounded below, has a least element. For example, there is no least
value among the positive real numbers. For any positive r, the number r/2 is still smaller,
and 0 cannot be counted since it is not positive, hence not included. The idea here is that
a set of upper bounds must include a member which plays the same role as 0 does for the
nonnegative real numbers.)
The full statement is that if (an ) is an increasing sequence which is bounded above,
then it is convergent and its limit is the least upper bound. Suppose (an ) has least upper
bound L, and let be a positive number. If we had an ≤ L − for all n, then L − would
be an upper bound which is less than the least upper bound. Since this cannot happen,
there must be an integer N for which aN > L − . Since (an ) is increasing, an > L − for
all n ≥ N . And since L is an upper bound, an < L + for all n. Together these inequalities
give |an − L| < for all n ≥ N , which insures that limn→∞ an = L.
Similarly, if (an ) has a lower bound, then it has a greatest lower bound, and if (an ) is
also decreasing, it must be convergent with this greatest lower bound as its limit.
Example 3.1. Consider the sequence defined recursively by
i) a0 = 0,
ii) an = 1/2(an−1 + 1), for n ≥ 1.
The first terms are a0 = 0, a1 = 1/2(0 + 1) = 1/2, and a2 = 1/2( 1/2 + 1) = 3/4.
We can show that this sequence is bounded above by B = 1 inductively. First a0 ≤ 1.
Then if an−1 ≤ 1, we have an−1 + 1 ≤ 2, so an = 1/2(an−1 + 1) < 1/2(2) = 1 as well.
We can show that the sequence is increasing by induction too. Clearly a1 = 1/2 > a0 .
If an > an−1 , then an + 1 > an−1 + 1 and 1/2(an + 1) > 1/2(an−1 + 1), which says that
an+1 > an .
Since this is an increasing sequence which is bounded above, it must converge. Its
limit is its least upper bound, whatever that is. This type of reasoning does not usually tell
us how to compute the limit – only that it must exist. In this case one might guess from
15
§3. Bounded Sequences
the first terms, 0, 1/2, 3/4, 7/8, 15/16, 31/32, . . ., that an = 1 − 1/2n for all n ≥ 0. That this is
really so can also be proved by induction, which makes it easy to see that limn→∞ an = 1.
Example 3.2. Consider the sequence defined by
i) a0 = 1
√
ii) an = 1 + an−1 , for each n ≥ 1.
Here the first few terms are a0 = 1, a1 =
p
√
√
√
1 + 1 = 2 ≈ 1.41, and a2 = 1 + 2 ≈ 1.55.
Again the sequence is bounded above. For example , we can show that an ≤ 2 for each
√
√
n ≥ 0 since a0 = 1 < 2, and if an−1 ≤ 2, then an = 1 + an−1 ≤ 1 + 2 ≈ 1.73 < 2. It is
√
also increasing. This too follows by induction. First a1 = 2 ≥ a0 . Then if an ≥ an−1 ,
√
√
we have 1 + an ≥ 1 + an−1 which insures that an+1 ≥ an as well.
We can conclude that this sequence converges and that its limit is at most the upper
bound 2. Since the sequence is increasing, the limit must also be at least as large as
a2 ≈ 1.55. The first terms of the sequence of Example 3.1 formed a pattern from which
it was possible to guess a general formula, an = 1 − 1/2n , which turned out to be correct,
and from which it was easy to find the limit. That is not the case here.
Eventually we will develop some better methods for finding a general formula for the
nth term of a sequence when it is given by a description such as that in Example 3.1.
These methods will usually not work, however, if the description includes a “nonlinear”
term such as a square, or as in the case of Example 3.2, a square root. In such cases,
although we may not be able to find a general formula for an , we will still seek ways to
analyze the long range behavior of the sequence.
For the moment, note that we could get better information about the limit of the
sequence in Example 3.2 by computing, more terms. Since (an ) is increasing, the limit
must be at least as large as a3 which is approximately 1.6. From the other direction we
could try to establish tighter upper bounds. An argument exactly like that used to show
that an ≤ 2 could be used to show that an ≤ 1.7 for all n ≥ 0. The limit must therefore
be in the interval [1.6, 1.7].
We do know that the limit exists, since the sequence is increasing and bounded above.
Call this limit L. For large n both an and an+1 are close to L. But the definition of the
16
§3. Bounded Sequences
sequence tells us that a2n+1 = 1 + an for all n ≥ 0. So L2 must be close to 1 + L. In
fact, the sequence (a2n+1 ) converges to L2 and the sequence (1 + an ) to 1 + L. So with
a2n+1 = 1 + an for n ≥ 0, we must have L2 = 1 + L. The limit must be one of the roots of
L2 − L − 1 = 0.
Since the smaller of the two roots is less than one, L must be the larger root which is
√
1+ 5
2 .
That is
√
1+ 5
lim an =
≈ 1.618.
n→∞
2
Example 3.3. Consider the sequence (an ) defined by
an = (1 + 1/n)n ,
for n ≥ 1.
The first few terms of this sequence are a1 = (1 + 1/1)1 = 2, a2 = (1 + 1/2)2 = 9/4 = 2.25,
a3 = (1 + 1/3)3 =
64/27
≈ 2.37, and a4 = (1 + 1/4)4 =
625/256
≈ 2.44.
These first terms suggest that the sequence is increasing. Checking a finite number of
terms is never enough, however, to insure that an+1 ≥ an for the entire sequence. There
is always the possibility that the behavior changes at some point beyond those which have
been considered. Here the expression 1 + 1/n actually decreases as n increases. But it is
never less than or equal to 1, and for n > 1 the nth power of a number larger than 1 is
larger than that number and grows as n increases. The combined effect of the decrease
in 1 + 1/n and increase of its exponent is not easy to analyze. The fact is that this is a
(strictly) increasing sequence, but a direct comparison of an and an+1 requires some work.
It can be done by using the binomial theorem to expand (1 + 1/n)n . This is outlined in
Exercise 3.2.
Since the first terms are growing rather slowly, one also suspects that this sequence
has an upper bound. Again we need to be careful. The sequence bn = (1.01)n also grows
very slowly. But it has no upper bound, and therefore no limit. For any number M > 0
we will have (1.01)n > M for all n which satisfy n ln 1.01 > ln M , where ln denotes the
natural logarithm. The value ln 1.01 is small, but it is positive, so (1.01)n > M for all
17
§3. Bounded Sequences
n > ln M/ ln 1.01. The same argument shows that cn = r n grows without bound for each
r > 1.
The sequence an = (1 + 1/n)n is bounded above, but this too cannot be checked too
easily. Exercise 3.3 contains an outline of a proof that 3 is an upper bound.
Since an = (1 + 1/n)n is increasing and bounded above, it must converge to some limit.
In order to find that limit we can consider the related sequence
bn = ln an = ln ((1 + 1/n)n ) = n ln(1 + 1/n).
If we can find the limit limn→∞ bn = L (which must exist by Theorem 2.2), then using
that theorem and the fact that the function f (x) = ex is continuous for all x, we will have
lim an = lim eln an = lim ebn = eL .
n→∞
n→∞
n→∞
Taking the logarithm allows us to work with a product rather than a power, but one term
of this product, n, grows without bound, while the other, ln(1 + 1/n), has limit 0 (since
(1 + 1/n) is approaching 1, and since g(x) = ln x is continuous for x > 0, ln(1 + 1/n) tends
to ln(1) = 0). It is impossible to predict the behavior of the product from this information
alone. In the products n · 1/n2 , n · 1/n, and n2 · 1/n the first term grows without bound,
while the second term has limit 0. But the first product has limit 0, the second limit 1,
and the third has no finite limit at all.
Our solution in this case is based on the calculus and the observation that
bn = n ln(1 + 1/n) =
ln(1 + 1/n) − 0
ln(1 + 1/n) − ln(1)
ln(1 + 1/n)
=
=
.
1/n
1/n
1/n
The right side is the value at 1/n of the quotient which is the basis of the definition of the
derivative of g(x) = ln x at x = 1. Its limit as n increases and 1/n tends to zero is the value
of that derivative at x = 1. This gives
lim bn =
n→∞
d
ln x|x=1 = 1/x|x=1 = 1.
dx
This in turn means that
lim bn
lim an = en→∞
n→∞
= e1 = e ≈ 2.7182.
18
§3. Bounded Sequences
The convergence of (1 + 1/n)n to the limit e is quite slow, so computing successive terms
of this sequence does produce a list of approximations to the irrational number e, but is
not a very efficient way of doing so. Eventually we will see that e is also the limit of the
sequence
an = 1 + 1/1! + 1/2! + 1/3! + · · · + 1/n!
which converges much more rapidly.
Exercises
3.1. Find an upper bound for the sequence, or show that there is none.
i) an = ( 3/2)n , for n ≥ 0
ii) a0 = 0, a1 = 1, and an+1 = 1/2(an + an−1 ), for n ≥ 1
iii) a0 = 0, a1 = 1, and an+1 = an + an−1 , for n ≥ 1
3.2. Show that an = (1 + 1/n)n is increasing by using the binomial theorem to expand
(1 + 1/n)n as
n
n
n
1
2
(1 + 1/n) = 1 +
( 1/n) +
( 1/n) + · · · +
( 1/n)n−1 + ( 1/n)n ,
1
2
n−1
n
then comparing the individual terms with the corresponding terms of the expansion
of an+1 = (1 + 1/n+1)n+1 .
3.3. Show that an = (1+ 1/n)n is bounded above by checking that in the binomial expansion
of an the k th term is at most 1/2k−2 for each k > 2.
3.4. Find the limits of the following sequences
i) an = (1 + 2/n)n for n ≥ 1
ii) an = (1 − 1/n)n for n ≥ 1
iii) an = (1 + 1/n2 )n for n ≥ 1
2
(Note, for example, that (1 + 2/n)n = (1 + 1/n/2)n/2 .)
3.5. The sequence (an ) is increasing if and only if an+1 /an ≥ 1 for each n. If
limn→∞ an+1 /an exists in this case, it must be least 1.
i) What can be concluded if limn→∞ an+1 /an > 1?
ii) Find an increasing sequence (an ) with limn→∞ an+1 /an = 1 which is convergent.
iii) Find an increasing sequence (an ) with limn→∞ an+1 /an = 1 which is divergent.
3.6. Show that if the sequence (an ) has a limit, then for each positive number there is
an integer N such that |an+1 − an | < for each n ≥ N .
19
§4. Rate of Growth
§4. Rate of Growth
We have been concentrating on convergent sequences and their limits, but in many
applications we will be dealing with a sequence whose nth term, an , is the size of some
quantity which increases without bound as n increases. The most common example of this
will be the case in which an is the number of operations of some particular type required
to solve a problem whose size by some measure is n when a certain algorithm is used. The
larger the input problem, the larger the number of steps required to solve it. We would
not expect a sequence of such values to converge.
In cases such as this our interest will not be in finding a limit, but rather in the rate at
which the sequence grows. To study this we will consider ways to compare the growth rates
of two sequences. Then we will assemble a list of increasing sequences which can serve as
“benchmarks” to which other sequences can be compared. In view of the applications, we
will restrict ourselves for the most part to sequences of positive terms, but the standard
definitions, since they are stated in terms of absolute values, will apply more generally, the
first of these involves notation which has been in use in mathematics for some time.
Definition 4.1. For sequences (an ) and (bn ) we say that
an = O(bn )
(read “an is big Oh of bn ”) provided that there is an integer N and a (nonnegative) constant
C such that
|an | ≤ C|bn |
for all n ≥ N . The idea behind this “big Oh” notation for increasing sequences (an ) and
(bn ) is that if an = O(bn ), then (an ) does not grow at a rate significantly greater than that
of (bn ). Except for possibly a finite number of terms (those with n < N ), |an | is never
larger than some constant multiple of |bn |.
Example 4.1. Consider the sequences (an ) and (bn ) with an = 2n + 5 and bn = 2n . We
claim that
2n + 5 = O(2n ).
20
§4. Rate of Growth
To verify this we need to show that there is an integer N and constant C such that for all
n ≥ N we have
2n + 5 ≤ C · 2 n .
This is equivalent to requiring that
2n + 5
≤C
2n
or
1+
5
≤ C,
2n
for sufficiently large n.
Since 2n ≥ 5 for n ≥ 3 we have 5/2n ≤ 1 and 1 + 5/2n ≤ 2 for such n, and we could
retrace our steps to show that
2n + 5 ≤ 2 · 2 n
for all n ≥ 3.
Note that since 2n ≤ 1(2n + 5) for all n ≥ 0 we also have
2n = O(2n + 5).
Example 4.2. Notice that the comparison on which an = O(bn ) depends concerns the
long term behavior of the two sequences, and may not be a good description of how they
begin. Thus we have
106 n2 + 10n + 1 = O(n3 ).
Again to show that there is a constant C such that 106 n2 +10n+1+ ≤ C ·n3 for sufficiently
large n we consider the ratio
106 n2 + 10n + 1
10
106
1
+ 2+ 3.
=
3
n
n
n
n
It is easy to see that for all n ≥ 106 each term on the right is at most 1, so for such n we
have
106 n2 + 10n + 1 ≤ 3 · n3 .
While we interpret this as meaning that 106 n2 + 10n + 1 grows no faster than n3 , note
that its values are larger than those of n3 for the first one million terms and are still more
than one hundred times as large for n = 104 .
21
§4. Rate of Growth
In this example we cannot reverse the order of the sequences. We do not have n 3 =
O(106 n2 + 10n + 1) since there is no C such that n3 ≤ C(106 n2 + 10n + 1) remains true
for all n beyond some point. To check this, notice that for any positive constant C we
have n3 > C(106 n2 + 10n + 1) for all n with n > 2 · C · 106 , since in this case we have
n3/2
n3/2
> C · 106 n2 , and consequently
> C(10n + 1) as well.
When, as in Example 4.1, we do have both an = O(bn ) and bn = O(an ), we will say
that the two sequences have the same rate of growth. This is often denoted by (a n ) = Θ(bn ).
This is equivalent to requiring that there exist positive constants c and C and an integer
N such that for all n ≥ N we have
c|an | ≤ |bn | ≤ C|an |.
For sequences of nonzero terms this can also be written as
c ≤ |bn /an | ≤ C,
or, since c and C must be positive, as
1/c
≤ |an /bn | ≤ 1/C
for all n ≥ N .
For the sequences an = 2n + 5 and bn = 2n of Example 4.1 the ratio |bn /an | =
2n/2n +5
= 1 − 5/2n +5 is easily seen to have limit 1. This can be used to show that the two
sequences have the same rate of growth. By considering = 1/2 we know that there is an
integer N such that for n ≥ N we have
n
| 2 /2n +5 − 1| < 1/2
and therefore that
1/2
n
≤ | 2 /2n +5| ≤ 3/2.
Whenever the limit limn→∞ |bn /an | = L exists and is positive we can find constants c and
C which are positive and for which c ≤ |bn /an | ≤ C for all sufficiently large n. We need
only consider a positive number such that L − > 0. (We could use = L/2 for example.)
Then we know that there is no N such that for n ≥ N we have
L − ≤ |bn /an | ≤ L + .
This provides a proof of the next statement.
22
§4. Rate of Growth
Theorem 4.1. If for sequences (an ) and (bn ) we have
lim |bn /an | = L 6= 0.
n→∞
Then (an ) and (bn ) have the same rate of growth. That is
an = Θ(bn ).
Note that our definition of (an ) = Θ(bn ) did not require that the limit limn→∞ |bn /an |
exists (see Exercise 4.2). But if there is such a limit, we have (an ) = Θ(bn ) exactly when
that limit is positive. Suppose, instead, as for the sequences of Example 4.2, we have a
ratio whose limit is zero.
6 2
6
10 n + 10n + 1
10
10
1
lim
lim
+ 2 + 3 = 0.
= n→∞
n→∞ n3
n
n
n
We cannot conclude that 106 n2 + 10n + 1 and n3 have the same rate of growth, but this
does insure that 106 n2 + 10n + 1 = O(n3 ) since for each positive , there is an integer N
such that
for n ≥ N , and therefore
6 2
10 n + 10n + 1
<
−
0
n3
106 n2 + 10n + 1 ≤ · n3 .
But on the other side we only have
−n3 < 106 n2 + 10n + 1.
In fact, since the limit is 0, we cannot have 106 n2 + 10n + 1 = Θ(n3 ). With limit 0 there
will be no positive constant c such that the ratio will remain as large as c for all n beyond
some point. (Since we could take = c, there must be an N such that the ratio is strictly
less than c for all n ≥ N .)
This illustrates the fact that if |bn /an | is convergent with limit 0, then bn = O(an ),
but we do not have bn = Θ(an ). The sequence (bn ) grows no faster than (an ) but is not
considered to have the same rate of growth as an . It is natural in this case to say that (bn )
has a smaller rate of growth than (an ). This is often denoted by “little Oh” notation as
bn = o(an ).
23
§4. Rate of Growth
Again there are examples (see Exercise 4.5) in which we have bn = O(an ) but not bn =
Θ(an ) when the sequence |bn /an | has no limit at all. The “little Oh” notation is usually
reserved, however, for the case when the limit limn→∞ |bn /an | does exist, and this case it
means exactly that this limit must be 0. We will adopt this interpretation. This may seem
somewhat inconsistent, but stems from the fact that there may be a constant C such that
|bn /an | ≤ C for all large n while there is no limit for |bn /an |. If for every positive C we
have |bn /an | ≤ C when n is large enough, then |bn /an | must converge to 0.
Exercises
4.1. Determine whether an = O(bn ), bn = O(an ), or both.
i) an = 2n2 − 3n, bn = n3
ii) an = 2n2 − 3n, bn = 10n2
iii) an = 2n , bn = 3n
iv) an = ( 1/2)n , bn = 1
4.2. Show that if an = 2n and bn = (2+sin n)n, then we have bn = Θ(an ), but that |bn /an |
has no limit.
4.3. Show that if bn = Θ(an ), then an = Θ(bn ).
4.4. Show that if bn = Θ(an ) and an = Θ(dn ), then bn = Θ(dn ).
4.5. Give an example of sequences (an ) and (bn ) for which an = O(bn ), we do not have
an = Θ(bn ), and for which |an /bn | has no limit.
4.6. Do there exist sequences (an ) and (bn ) such that an = o(bn ) and bn = o(an )? If so,
give an example.
4.7. Show that an = o(bn ), where
i) an = (1 + 1/n)n , bn = n
ii) an = 1/n − 1/n+1, bn = 1.
24
§5. Some Standard Sequences
§5. Some Standard Sequences
The rates of growth of most sequences which arise in applications are usually described
by comparing them with the behavior of certain “standard” sequences. For this reason
it is useful to consider the relationships among the members of such a list of benchmark
sequences. This section is devoted to such comparisons.
1. If an is given by some polynomial in n, then it is easy to compare the rate of growth
of an with sequences of the form bn = nj , where j is a nonnegative integer. Specifically, if
for some nonnegative integer k
an = ck nk + ck−1 nk−1 + · · · + c1 n1 + c0 ,
where ck > 0, then we have
an = Θ(nk ),
and
an = o(nj ),
for each j > k.
This follows from the facts that
lim
n→∞
an
ck−1
c1
c0 = ck
=
lim
c
+
+
·
·
·
+
+
k
n→∞
nk
n
nk−1
nk
and for j > k,
lim
n→∞
c
an
c0 k
= 0.
=
lim
+
·
·
·
+
n→∞ nj−k
nj
nj
For the sequences of Example 4.1 of the last section we have
106 n2 + 10n + 1 = Θ(n2 )
and
106 n2 + 10n + 1 = o(n3 ).
2. Comparing positive powers of n other than integers is just as easy. If r and s are
positive real numbers with r < s, then we have
nr = o(ns ).
25
§5. Some Standard Sequences
This follows immediately from the fact that with 0 < r < s we have
lim
n→∞
1
nr
=
lim
= 0.
ns n→∞ ns−r
Thus, for example, we have
√
(n) = o(n3/2 ).
n = o(n) and
3. Suppose r and s are positive real numbers with r < s. Then we have
r n = o(sn ).
In this case we have
lim
n→∞
rn
= lim ( r/s)n = 0,
n→∞
sn
since | r/s| < 1.
This gives, for example, that (2n ) = o(3n ), and similarly, (5 · 2n + 1) = o(3n ), since
limn→∞
5·2n +1
3n
= limn→∞ (5( 2/3)n + 1/3n ) = 5 · 0 + 0 = 0.
4. The next relationship has many applications.
log2 n = o(n).
Here log2 n denotes the logarithm to the base 2. That is, log2 n = r if n = 2r . Actually
an = log2 n grows much more slowly than bn = n. (Think of n = 2j for which an = j and
bn = 2j . With j = 20, for example, we have an = 20 and bn = 1, 048, 576.) A proof of this
fact still requires some work. Since we will base several other results on this one, we will
sketch a proof of it here. First notice that a similar result holds for any logarithm with
some base b, b > 1. This is because we have
logb n =
1
ln n,
ln b
which follows from the fact that
eln n = n = blogb n = (eln b )logb n = e(ln b)(logb n) .
26
§5. Some Standard Sequences
Thus for each b > 1, the logarithm logb n is a constant multiple of the natural logarithm
ln n. It follows that we need only show that
ln n = o(n).
Recall from the calculus that we have
ln n =
Z
n
1/t dt.
1
The area under the curve f (t) = 1/t represented by this integral can be bounded above by
a sum of areas of rectangles to give
ln n < 1 + 1/2 + · · · + 1/n−1
for each n ≥ 2.
1
1
1/2
2
3
n−1
n
It follows that we have
ln n
1 + 1/2 + · · · + 1/n−1
≤
n
n
Now to show that (ln n) = o(n) it is enough to check that limn→∞
(1+ 1/2+···+ 1/n−1)
n
= 0.
Some care is needed here since the number of terms in the numerator grows as n increases.
The proof can be completed, however, by showing that
1 + 1/2 + · · · + 1/n−1
2
≤ √
n
n
for all n ≥ 1, which can be proved by using induction to show (see Exercise 5.11)
√
1 + 1/2 + · · · + 1/n−1 ≤ 2 n.
27
§5. Some Standard Sequences
5. A number of relationships follow easily from the fact that limn→∞
log2 n
n
= 0. A
few of these are the following:
i) log2 n = o(nj ) for each positive integer j.
ii) log2 (nj ) = o(n) for each positive integer j.
iii) nj log2 n = o(nj+1 ) for each positive integer j.
The first of these follows from the fact that
log2 n
nj
≤
log2 n
n
for any integer j ≥ 1, the second
from the fact that log2 (nj ) = j log2 n, and the third from the observation that
log2 n
nj log2 n
=
lim
= 0.
n→∞
n→∞
nj+1
n
lim
The logarithm to the base 2 could be replaced in each of these statements by the logarithm
to any base b, b > 1, since logb n =
ln 2
ln b
log2 n.
The first of these results allows us to conclude that the sequence (log 2 n) grows more
slowly than any sequence (an ), where an is given by the value of some polynomial function
of n in which the coefficient of term of highest degree is positive. The next result tells us
that the sequences given by polynomials in n, in turn, grow more slowly than those given
by exponential functions of n.
6. For any positive integer j and any real number r, r > 1, we have
nj = o(r n ).
In order to show that limn→∞
nj/rn
= 0 we make use of the version of part (ii) of the
previous result for the logarithm to the base r. Since j + 1 is a positive integer we have
lim
n→∞
logr (nj+1 )
= 0.
n
It follows that for = 1 there exists an integer N such that for all n ≥ N we have
and hence that
logr (nj+1 )
− 0 < 1,
n
logr (nj+1 ) < n
28
§5. Some Standard Sequences
for all such n. Since r > 1, the function f (x) = r x is increasing, so by taking powers of r
we have
nj+1 = r log r
(nj +1)
< rn ,
and therefore that
1
nj
<
n
r
n
for all n ≥ N . The result now follows from the fact that limn→∞ 1/n = 0.
We now have that among the increasing sequences which arise most often in applications, those given by logarithms grow more slowly than those given by polynomials of n,
and these in turn grow more slowly than exponential expressions. Our last statements of
this type involve two more sequences which are important in many examples, and which
grow even more rapidly.
7. For each positive real number r we have
r n = o(n!).
Here n! denotes the factorial function n! = 1 · 2 · 3 · . . . · n. We illustrate the proof of this
statement by considering the case of r = 2, in which we must show that
2n
lim
= 0.
n→∞ n!
This follows immediately from the observation that
2n/n!
= 2/1 · 2/2 · ( 2/3 · · · 2/n−1) · 2/n ≤ 2/1 · 2/2 · 2/n = 4/n
for all n ≥ 3, since the term in parentheses is a product of terms, each at most one. Now
we only need to note that limn→∞ 4/n = 0.
Since n! is a product of n terms ranging from 1 to n, we can get a crude approximation
to its value by replacing each term in this product by the average of all of them. That
average is
n−1/2.
Using n/2 instead gives
n! ≈ ( n/2)n =
29
nn/2n .
§5. Some Standard Sequences
This rough approximation gives some idea about the rate of growth of n!. It can be shown
that a much better approximation to n! involves ( n/e)n instead of ( n/2)n , where e ≈ 2.7182
is the base of the natural logarithm. This very good approximation is known as Stirling’s
formula
n! ≈
√
2πn( n/e)n .
In this expression the ratio of the left side over the right has limit 1 as n increases, so in
√
the notation of the previous section n! = Θ 2πn( n/e)n .
Our last statement could be based on this approximation of n!, but it can also be
shown directly quite easily.
n! = o(nn )
8.
To show that limn→∞ n!/nn = 0 we can again write the ratio as a product and note that
for n ≥ 2, we have
n!
= 1/n · ( 2/n · 3/n · . . . · n/n) ≤ 1 · 1/n
nn
since each term in the parentheses is at most 1. The result follows from the fact that
limn→∞ 1/n = 0.
Exercises
5.1. Place the following sequences in an order such that each one grows more slowly than
those after it. That is, if (an ) comes before (bn ), then an = o(bn ).
i) n log n,
ii) n3 ,
iii) n2 − log n,
iv) 2n ,
v) n2 / log n.
√
5.2. Show that log2 n = o( n).
5.3. Show that (log2 n)j = o(n) for each positive integer j.
5.4. Show that if an = o(bn ) and bn = o(cn ), then an = o(cn ).
3
5.5. The statement n3 = o(2n ) is based on the fact that limn→∞ n /2n = 0. This insures
3
that for each > 0, there is an integer N such that n /2n < for all n ≥ N . For = 1
what is the smallest such integer N . Repeat for = 1/10.
5.6. Show that log2 log2 n = o(log2 n), where the left side is the logarithm of log2 n.
5.7. Show that 3n = o(n!).
30
§5. Some Standard Sequences
5.8. Show that n · 2n = o(3n ).
2
5.9. Show that 3n = o(2n ).
5.10. Give an example of a sequence (an ) such that nn = o(an ).
5.11. Show that for all n > 1 we have
√
1 + 1/2 + 1/3 + · · · + 1/n−1 ≤ 2 n.
31
§6. Finite Differences
§6. Finite Differences
The very first sequence we encountered began with the terms
6, 7, 9, 13, 21, 37, 71, 133.
This certainly suggests that the sequence is increasing, but to compare its rate of increase
carefully with that of another sequence, we would really need a general description of the
nth term an . Although these eight terms may not suggest a pattern by themselves, suppose
we consider their differences, that is terms of the form
an+1 − an .
In this example we get
7 − 6, 9 − 7, 13 − 9, 21 − 13, 37 − 21, 71 − 37, 133 − 71
or
1,
2,
4,
8,
16,
32,
64.
These first differences fit a very clear pattern, namely an+1 − an = 2n .
In the sequence which begins with the terms
1, 4, 7, 10, 13, 16, 19, 22,
the sequence of differences an+1 − an is even simpler. It is
3, 3, 3, 3, 3, 3, 3.
Very often we can obtain information about the original sequence by considering such a
sequence of its differences, or even by taking differences of the terms of this new sequence.
We are about to see how this can be done. To do so, it is convenient to give a formal name
to the operation of replacing a sequence by the new sequence of its differences. By the
forward difference operator, ∆, we mean the function which assigns to each sequence (a n )
the sequence of its differences (an+1 − an ). We will write this as
∆(an ) = an+1 − an .
Example 6.1. If an = 2n + 1 for each n ≥ 0, then
∆(an ) = ∆(2n + 1) = (2(n + 1) + 1) − (2n + 1) = 2
The result in this case is the constant sequence each of whose terms is 2.
32
§6. Finite Differences
Example 6.2. If (an ) is itself a constant sequence with an = c for each n, we obtain
∆(an ) = ∆(c) = c − c = 0.
Now the result is the sequence of all zeros.
Example 6.3. For an = n2 , n ≥ 0, we have
∆(an ) = ∆(n2 ) = (n + 1)2 − n2 = 2n + 1.
The result is the original sequence of Example 6.1.
Example 6.4. Consider the exponential sequence an = 3n . Here the difference operator
gives
∆(an ) = ∆(3n ) = 3n+1 − 3n = (3 − 1) · 3n = 2 · 3n .
The differences form a sequence whose terms can be obtained by multiplying the terms of
the original sequence by 2. This could be written as
∆(an ) = 2 · an .
Suppose we form a sequence by adding corresponding terms of the sequences in Examples 6.3 and 6.4. That is, let an = 3n + n2 . Applying the difference operator yields
∆(3n + n2 ) = (3n+1 + (n + 1)2 ) − (3n + n2 )
= (3n+1 − 3n ) + ((n + 1)2 − n2 ) = 2 · 3n + (2n + 1) = ∆(3n ) + ∆(n2 ).
The difference of the sum is the sum of the differences.
The difference of any sum can be computed this way, one part at a time, since
∆(an + bn ) = (an+1 + bn+1 ) − (an + bn ) = (an+1 − an ) + (bn+1 − bn ) = ∆(an ) + ∆(bn ).
There is also a simple rule for computing the differences of a sequence formed by multiplying
each term of another sequence by the same constant. Consider, for example, the sequence
an = 5n2 . Taking differences give
∆(5n2 ) = 5(n + 1)2 − 5n2 = 5((n + 1)2 − n2 ) = 5(2n + 1) = 5∆(n2 ).
The constant coefficient is simply carried along.
Again, this is true in general, since for any sequence (an ) and constant c we have
∆(can ) = can+1 − can = c(an+1 − an ) = c∆(an ).
Combining these facts we have the following.
33
§6. Finite Differences
Theorem 6.1. For any sequences (an ) and (bn ) and any constants c and d we have
∆(c · an + d · bn ) = c∆(an ) + d∆(bn ).
Any function with this property is called linear. Thus ∆ is an example of a linear operator.
(The term “operator” is used in many circumstances in place of the word “function,” as
is the word “transformation.”)
Some of the most familiar linear functions are the linear transformations from a vector
space to itself or to another vector space defined by matrix multiplication. If A is an m × n
matrix, then the function from vectors of length n to vectors of length m given by
T (x) = Ax
is linear since for vectors x and y and constants c and d we have
T (cx + dy) = A(cx + dy) = cAx + dAy = cT (x) + dT (y).
There is another familiar linear operator, although you may have never thought of it
as such. Rather than sequences or vectors, its domain consists of functions. And its values
are functions. This is differentiation. If we use operator notation, and let D denote the
differential operator defined for differentiable functions y = f (x) by
D(f (x)) =
df
= f 0 (x),
dx
then for functions f and g and constants c and d we have
D(cf + dg) = (cf + dg)0 = cf 0 + dg 0 = cD(f ) + dD(g).
There are many strong parallels between the difference operator ∆ and the differential
operator D, beyond the fact that both are linear. The difference operator could also be
defined for functions as well as sequences. Recall that each sequence (an ) may be viewed
as the list of values of some function f at the integers 0, 1, 2, . . . , n, . . . , with an = f (n).
Viewed this way, the difference operator gives
∆(an ) = an+1 − an = f (n + 1) − f (n).
34
§6. Finite Differences
This suggests the following definition of ∆ as an operator whose domain, like that of the
differential operator D, consists of functions, and whose values are also function. For each
function f (x) we define the function ∆(f (x)) by
∆(f (x)) = f (x + 1) − f (x).
Technically, this is a different operator since its domain consists of functions rather than
sequences. But these two uses of ∆ are consistent in the sense that if (an ) is the list of
values of the function f at 0, 1, 2, . . ., then the sequence ∆(an ) consists of the values of the
function ∆(f ) at the nonnegative integers. We will therefore use the same notation for
these two difference operators, applied to sequences and applied to functions.
Example 6.5. For the function f (x) = x2 we have
∆(x2 ) = (x + 1)2 − x = 2x + 1.
Restricting these functions to the nonnegative integers gives the statement about sequences
from Example 6.3, ∆(n2 ) = 2n + 1.
The very same reasoning used to establish Theorem 6.1, with some simple changes of
notation now gives
Theorem 6.2. The forward difference operator ∆ as applied to functions is a linear
operator. That is, for any functions f and g and any constants c and d
∆(c · f + d · g) = c · ∆(f ) + d · ∆(g).
An immediate consequence of the fact that the differential operator D(f ) = f 0 is
linear that we can find the derivative of any polynomial function as soon as we have
the derivatives of the integer powers of x (and of constants thought of as c · x0 ). Since
D(xj ) = jxj−1 for each integer j, we have
D(ck xk + ck−1 xk−1 + · · · + c1 x + c0 ) = ck · k · xk−1 + ck−1 (k − 1)xk−2 + · · · + c1 · 1 + 0.
The same reasoning for the difference operator applied to a polynomial function or to a
sequence (an ) where the nth term an is given by a polynomial in n would be
∆(ck xk + ck−1 xk−1 + · · · + c1 x + c0 ) = ck ∆(xk ) + ck−1 ∆(xk−1 ) + · · · + c1 ∆(x) + c0 ∆(1)
35
§6. Finite Differences
or
∆(ck nk + ck−1 nk−1 + · · · + c1 n + c0 ) = ck ∆(nk ) + ck−1 ∆(nk−1 ) + · · · + c1 ∆(n) + c0 ∆(1).
To make use of this we need to find ∆(xk ) for positive integers k, or, for sequences, ∆(nk ).
For constant functions and sequences we have (see Example 6.2)
∆(c) = 0.
Next, it is easy to see that
∆(x1 ) = (x + 1)1 − x1 = 1
and
∆(n1 ) = (n + 1)1 − n1 = 1.
But by Examples 6.3 and 6.5 we have
∆(x2 ) = 2x + 1 and
∆(n2 ) = 2n + 1.
And direct computation gives
∆(x3 ) = (x + 1)3 − x3 = (x3 + 3x2 + 3x + 1) − x3 = 3x2 + 3x + 1,
so
∆(n3 ) = 3n2 + 3n + 1.
We can already see that – unfortunately – the value of ∆ at a power of x (or of n) is not
as simple as the value of the derivative at such a power. It is not just a matter of using
the original exponent as a coefficient and reducing the power by one. But, at least so far,
we do have something. In each of these cases the value of ∆(xj ) or ∆(nj ) is a polynomial,
and one of degree j − 1, when j ≥ 1. This much is true in general. As for the examples
of ∆(x2 ) and ∆(x3 ), when we expand (x + 1)j by multiplying it out or using the binomial
theorem, we obtain a sum of terms beginning with 1 · xj . When we take the difference
∆(xj ) = (x + 1)j − xj , this first term of (x + 1)j is cancelled by the second part of the
expression, and result is a polynomial of degree j − 1, beginning with 1j xj−1 = jxj−1 ,
having j terms in all.
j
j j−1
1
x + 1 − xj
x
+···+
∆(x ) = (x + 1) − x = x +
j−1
1
j j−2
j−1
x
+ · · · + jx + 1.
= jx
+
2
j
j
j
j
36
§6. Finite Differences
When we apply the difference operator to a polynomial of degree k, k ≥ 1, the result is
a sum of terms of the form cj ∆(xj ), 0 ≤ j ≤ k, where ck ∆(xk ) is a polynomial of degree
k − 1, and the rest are polynomials of lower degree. This gives a proof of the following
statement.
Theorem 6.3. If f (x) is given by a polynomial of degree k, k ≥ 1, then ∆(f (x)) is a
polynomial of degree k − 1.
We began this section by noting that it might be possible to find a formula for the
nth term, an , of a sequence (an ) by considering the sequence of differences ∆(an ). That
is, if the sequence ∆(an ) is simple enough, we might be able to work backwards to get
a description of the sequence which has these differences. This is analogous to having
the derivative of a function, the value of the differential operator D(f ) = f 0 , and seeking
the original function f . In other words, it is analogous to finding an indefinite integral.
R
We know that x2 dx = 1/3x3 + c since we know the derivative of x3 , and we obtain
R
cos xdx = sin x + 1 from the fact that D(sin x) = cos x. (We are also using the fact
that two functions which differ by a constant have the same derivative – note that they
also have the same difference.) In order to find integrals we need to know a collection
of derivatives. Similarly, in order to find “antidifferences,” we need some practice finding
differences. We have some examples already, and our next task is to add more.
Exercises
6.1. Find ∆(an ), where
i) an = 3n4 − 2n
ii) an = 7 · 2n
iii) an = n(n − 1)
iv) an = n · 2n
v) an = n!
6.2. Find the “second difference” ∆(∆(3n )).
6.3. Show that if f (x) = g(x) + c for some constant c, then ∆(f ) = ∆(g).
6.4. Show that if ∆(an ) = 0 for all n, then (an ) is a constant sequence. That is, an = c
for all n.
37
§6. Finite Differences
6.5. Use 6.4 to show that if ∆(an ) = ∆(bn ) for all n, then there is a constant c such that
an = bn + c for all n.
6.6. If ∆(f (x)) = ∆(g(x)), does it follow that f (x) = g(x) + c for some constant c and all
x?
6.7. The backward difference operator, ∇, is defined by ∇(f (x)) = f (x) − f (x − 1). Find
i) ∇(x2 )
ii) ∇(3n ).
6.8. The shift operator, E, is defined by E(f (x)) = f (x + 1) and the identity operator, I,
by I(f (x)) = f (x). Show that ∆ = E − I, meaning that for each f (x) we have
∆(f (x)) = E(f (x)) − I(f (x)).
38
§7. Antidifferences
§7. Antidifferences
We have begun studying properties of the difference operator ∆ so that we can deal
with questions of the form: Find those sequences (an ) for which ∆(an ) = f (n). The
following problem illustrates how this might be done.
Problem. Suppose the sequence (an ) is defined recursively by i) a0 = 5, and ii) an+1 =
an + 3n for each n ≥ 0. Find an .
The second part of the definition tells us that
∆(an ) = an+1 − an = 3n
for all n ≥ 0. Our approach is therefore to try to find all sequences with ∆(an ) = 3n , and
then find among them the sequence which begins with a0 = 5. (Notice that there must be
precisely one such sequence since the recursive definition yields a unique sequence – one
which begins with a0 = 5, a1 = 5 + 30 = 6, a2 = 6 + 31 = 9, etc.)
Since the differences are given by a formula which is exponential in n, we may guess
that the expression for an also involves an exponential term. For such terms we have the
following fact:
For each real number r
∆(r n ) = r n+1 − r n = (r − 1)r n .
The difference of an = r n is always just a multiple of itself. Thus, for example,
∆(3n ) = (3 − 1) · 3n = 2 · 3n ,
∆(( 1/2)n ) = ( 1/2 − 1)( 1/2)n = (− 1/2)( 1/2)n ,
and
∆(2n ) = (2 − 1)(2n ) = 1 · 2n .
There is exactly one case, an = 2n , in which the coefficient in ∆(an ) is 1. That is, one
case in which we have ∆(an ) = an .
39
§7. Antidifferences
(Here we have another similarity to properties of the differential operator D(f ) = f 0 .
When we apply this operator to an exponential function, that is when we differentiate a
function of the form f (x) = bx , b > 0, we obtain
D(bx ) = D((eln b )x ) = D(e(ln b)x ) = ln b · eln(b)x = ln b · bx .
The derivative of f (x) = bx is a multiple of bx , and there is exactly one case where the
coefficient is one, namely when ln b = 1, which occurs when b = e. In some sense then, the
sequence an = 2n , or function f (x) = 2x , plays a role for ∆ corresponding to that of the
exponential function f (x) = ex for differentiation.)
Coming back to the original problem, in which ∆(an ) = 3n , we now have an expression
which will at least produce a difference which contains the term 3n , since we have ∆(3n ) =
2 · 3n . It is easy to see what to do now. By the linearity of ∆ we have ∆(c · an ) = c∆(an ),
so
∆( 1/2 · 3n ) = 1/2∆(3n ) = 1/2 · 2 · 3n = 3n .
The sequence an = 1/2 · 3n does have the required differences. For the sequence we seek,
however, we need a0 = 5 and here we have a0 = 1/2 · 30 = 1/2. There must be other
sequences with the same differences. (We know that the recursive definition does yield a
sequence with a0 = 5 and ∆(an ) = 3n for all n ≥ 0.)
Again it is not hard to find more sequences with the same differences as an = 1/2 · 3n .
We know that two sequences whose terms differ only by a constant will have the same
differences, just as two functions which differ by a constant have the same derivative.
That is, for any constant c,
∆(an + c) = ∆(an ) + ∆(c) = ∆(an ) + 0 = ∆(an ).
In fact, we know (see Exercise 6.5) that if two sequences do have the same differences, then
they must differ by an additive constant. It follows that the sequence we seek must be of
the form an = 1/2 · 3n + c, for some constant c. All we need to do now is find c. To do this
we can use the “initial condition,” a0 = 5. The sequence we seek must satisfy
5 = a0 = 1/2 · 30 + c = 1/2 + c,
40
§7. Antidifferences
so we have
c = 9/2,
and
an = 1/2 · 3n + 9/2.
(As a partial check on our arithmetic, we can use this formula to recompute the first few
terms, obtaining a0 = 5, a1 = 1/2 · 3 + 9/2 = 6, and a2 = 1/2 · 32 + 9/2 =
18/2
= 9. While
this provides no guarantee of our formula in general, it is an easy way to uncover errors.)
To carry out arguments such as this one, it is convenient to have notation for the
family of sequences which have a given difference sequence, just as the indefinite integral
represents the family of functions with a given derivative.
Definition 7.1. For a sequence (an ), the antidifference of (an ), denoted by ∆−1 (an ), is
the family of all sequences (bn ) for which ∆(bn ) = an .
As we have seen, if we know one sequence (bn ) in the family ∆−1 (an ), then we know
all such sequences, since every other is of the form (bn + c) for some constant c. We will
indicate this by writing
∆−1 (an ) = bn + c if and only if ∆(bn ) = an .
Using this notation in the last problem, we would have reasoned that, since ∆(an ) = 3n
and ∆(3n ) = (3 − 1) · 3n = 2 · 3n , then
an = ∆−1 (3n ) =
1
· 3n + c = 1/2 · 3n + c,
3−1
and that since a0 = 5, for this problem we have c = 9/2.
From the examples we have considered so far we have
(1) For any r, r 6= 1, we have
∆(r n ) = (r − 1)r n , therefore ∆−1 (r n ) =
(2) For any constant c we have
∆(c), therefore ∆−1 (0) = c.
41
1
rn .
r−1
§7. Antidifferences
(3) ∆(n) = 1, therefore ∆−1 (1) = n + c.
(4) ∆(n2 ) = 2n + 1, therefore ∆−1 (2n + 1) = n2 + c.
The linearity of the difference operator ∆ also insures that
∆−1 (an + bn ) = ∆−1 (an ) + ∆−1 (bn ),
(5)
and for any constant d,
∆−1 (d · an ) = d∆−1 (an ).
In particular, if ∆(an ) is given by a polynomial in n,
∆(an ) = ck nk + ck−1 nk−1 + · · · + c1 n + c0 ,
then we have
an = ck ∆−1 (nk ) + ck−1 ∆−1 (nk−1 ) + · · · + c1 ∆−1 (n) + c0 ∆−1 (1).
This last fact should help to find (an ) when ∆(an ) is given by a polynomial in n, but there
is a problem. Since ∆(nj ) is not just jnj−1 , but a polynomial in n which begins with this
term, it is not clear how to find ∆−1 (nj ) for j > 0. Consider the case of j = 1. To find
∆−1 (n) we need to find one sequence (an ) with ∆(an ) = n. (All others will then differ
from this one by a constant term.) Using the facts that ∆(n2 ) = 2n + 1 and ∆(n) = 1, we
have
∆(n2 − n) = ∆(n2 ) − ∆(n) = (2n + 1) − (1) = 2n,
and therefore that
∆−1 (2n) = n2 − n + c.
Once we have this, it is easy to see that dividing by 2 gives
∆−1 (n) = ∆−1 ( 1/2 · 2n) = 1/2(n2 − n) + c =
n(n − 1)
+ c.
2
In the same way we could use ∆(n3 ) = 3n2 +3n+1, together with ∆−1 (n) = 1/2(n2 −n)+c
and ∆−1 (1) = n + c, to find ∆−1 (n2 ) as follows:
∆(n3 − 3 · 1/2(n2 − n)) = ∆(n3 ) − 3∆( 1/2(n2 − n)) = (3n2 + 3n + 1) − (3n) = 3n2 + 1,
42
§7. Antidifferences
so
∆(n3 − 3/2(n2 − n) − n) = (3n2 + 1) − ∆(n) = (3n2 + 1) − 1 = 3n2 .
Dividing by 3 yields
∆−1 (n2 ) = 1/3(n3 − 3/2(n2 − n) − n) = 1/3n3 − 1/2n2 + 1/6n =
(2n − 1)(n)(n − 1)
+ c.
6
To find ∆−1 (n3 ) we could use the fact that ∆(n4 ) = 4n3 +p(n), where p(n) is the quadratic
polynomial p(n) = 6n2 + 4n + 1. By subtracting the third degree polynomial ∆−1 (p(n)) =
q(n) we would obtain
∆(n4 − q(n)) = (4n3 + p(n)) − p(n) = 4n3 ,
and hence that
∆−1 (n3 ) = 1/4n4 = 1/4q(n),
where q(n) = ∆−1 (6n2 + 4n + 1) could be found using the previous examples.
We have the model here for an inductive proof that for each integer j, j ≥ 0, ∆−1 (nj )
is a polynomial of degree j + 1, whose first term is 1/j+1nj+1 . This fact, together with the
linearity of the operator ∆−1 , gives the following:
Theorem 7.1. If p(n) is a polynomial of degree k, then ∆−1 (p(n)) = q(n) + c, where q(n)
is a polynomial of degree k + 1.
Example 7.1. Suppose the sequence (an ) is given by
an = 1 + 2 + 3 + · · · + n,
for each n ≥ 1.
To find a compact formula for an we can consider the difference
∆(an ) = an+1 − an = (1 + 2 + · · · + (n + 1)) − (1 + 2 + · · · + n) = n + 1.
Since ∆(an ) is a polynomial of degree 1, we know by Theorem 7.1 that an is given by a
polynomial of degree 2. Here we have
an = ∆−1 (n + 1) = ∆−1 (n) + ∆−1 (1) = 1/2(n2 − n) + n + c = 1/2n2 + 1/2n + c,
43
§7. Antidifferences
for some constant c. Using the first term of the sequence gives
1 = a1 = 1/2(1)2 + 1/2(1) + c = 1 + c,
which shows that c = 0, so we have the formula
an = 1 + 2 + · · · + n = 1/2n2 + 1/2n =
n(n + 1)
.
2
Example 7.2. Let (an ) be the sequence defined by
i) a0 = 1
ii) an+1 = an + 3n + 1, for each n ≥ 0.
Since we have the difference ∆(an ) = an+1 − an = 3n + 1, we know that an is given
by a quadratic polynomial of the form an = An2 + Bn + C. We could use ∆−1 (n) and
∆−1 (1), as in the previous example to find the coefficients A and B. Alternatively, we
could use the values a0 = 1, a1 = 1 + 3 · 1 + 1 = 5, and a2 = 5 + 3 · 2 + 1 = 12 to find all
of the coefficients in this polynomial as follows:
1 = a0 = A · 02 + B · 0 + C = C
5 = a1 = A · 12 + B · 1 + C = A + B + C,
and
12 = a2 = A · 22 + B · 2 + C = 4A + 2B + C.
The first equation gives C = 1. Substituting this value in the other equations gives
A + B = 4 and 4A + 2B = 11 which have the solution A = 3/2, B = 5/2. It follows that
an = 3/2n2 + 5/2n + 1,
Exercises
7.1. Find ∆−1 (an ), where
i) an = 3n
ii) an = n2 + 2
iii) an = 4n
iv) an = n(n − 1).
44
for all n ≥ 0.
§7. Antidifferences
7.2. Find an for each of the following sequences.
i) a0 = 1 and an+1 = an + n for each n ≥ 0
ii) a0 = 2 and an+1 = an + 3n for each n ≥ 0
iii) a0 = 2 and an+1 = an + 3n+1 for each n ≥ 0
7.3. Suppose ∆(∆(an )) = 1 for all n ≥ 0, with a0 = 2. Find an .
7.4. Suppose (an ) is a sequence whose first six terms are
3, 4, 7, 12,
19, 28.
i) Find the first five terms of the sequence ∆(an ).
ii) Find a polynomial of n which gives the first five terms of ∆(an ).
iii) Find a polynomial of n which gives the first six terms of an .
7.5. Find ∆−1 (0) and ∆−1 (∆−1 (0)).
7.6. Show that ∆(n(n − 1)(n − 2)) = 3n(n − 1).
7.7. Find a sequence (an ) for which ∆(an ) = 3an for all n ≥ 0.
7.8. Find ∆−1 (n2n ). (First consider ∆(2n ) and ∆(n2n ).)
45
§8. Falling Factorials
§8. Falling Factorials
We have seen that for each integer j, j ≥ 1, the difference ∆(nj ) is a polynomial of
degree j − 1, and therefore, if p(n) is a polynomial of degree k, ∆(p(n)) is a polynomial of
degree k − 1.
In the other direction, for a polynomial p(n) of degree k, ∆−1 (p(n)) is a polynomial
of degree k + 1, since for each j, j ≥ 0, ∆−1 (nj ) is a polynomial of degree j + 1. The first
term of ∆−1 (nj ) is 1/j+1nj+1 , but there are no simple formulae which will produce the
other coefficients in the general case. As a result, they are not very useful in computing
the antidifference of a polynomial.
This is not like the case of the differential operator D and its inverse which gives
antiderivatives or indefinite integrals. The derivatives and integrals of powers, x j , are easy
to compute in general, and once we have these, we can handle all polynomials.
There exists, however, another family of sequences or functions which play a role for
differences and antidifferences similar to that of powers for derivatives and antiderivatives.
The basic idea can be seen from the following examples.
∆(n) = (n + 1) − n = 1
∆(n(n − 1)) = (n + 1)n − n(n − 1) = ((n + 1) − (n − 1))n = 2n
∆(n(n − 1)(n − 2)) = (n + 1)n(n − 1) − n(n − 1)(n − 2)
= [(n + 1) − (n − 2)]n(n − 1) = 3n(n − 1)
∆(n(n − 1)(n − 2)(n − 3)) = (n + 1)n(n − 1)(n − 2) − n(n − 1)(n − 2)(n − 3)
= [(n + 1) − (n − 3)]n(n − 1)(n − 2) = 4n(n − 1)(n − 2).
It is not hard to spot a pattern here. The basic expressions involved resemble factorials
in the sense that they are products of successive integers. For this reason they are called
falling factorials. They are defined by
n(k) = n(n − 1) · · · (n − k + 1).
Note that n(k) has exactly k terms and ends with n − k + 1 rather than n − k (since it
begins with n − 0). The corresponding functions are
x(k) = x(x − 1) · · · (x − k + 1).
46
§8. Falling Factorials
Thus, for example,
x(4) = x(x − 1)(x − 2)(x − 3).
Warning: There are other notations for falling factorials which are used at least as often
as this one. The most common are (x)k and xk .
The differences of falling factorials are easy to compute since we have
∆(n(k) ) = (n + 1)(n)(n − 1) · · · (n − k + 2) − n(n − 1) · · · (n − k + 1)
= [(n + 1) − (n − k + 1)]n(n − 1) · · · (n − k + 2)
= kn(n − 1) · · · (n − k + 2) = kn(k−1) .
For functions this becomes
∆(x(k) ) = kx(k−1) ,
for each k ≥ 1.
We can include the case of k = 1 here by defining
n(0) = x(0) ≡ 1,
so that
∆(n(1) ) = ∆(n) = 1 = 1n(0) .
The falling factorial n(k) can also be written as n!/(n − k)! and so is related to the binomial
coefficient nk by
n(k)
n!
n
.
=
=
k!
k!(n − k)!
k
The difference equation
∆(n(k) ) = kn(k−1) ,
k≥1
resembles the derivative formula
D(xk ) = kxk−1 ,
and, like it, leads immediately to a formula in the opposite direction.
∆−1 (n(k) ) =
1
n(k+1) ,
k+1
47
k ≥ 0.
§8. Falling Factorials
Thus, for example,
∆−1 (n) = ∆−1 (n(1) ) = 1/2n(2) = 1/2n(n − 1) = 1/2n2 − 1/2n.
∆−1 (n(n − 1)) = ∆−1 (n(2) ) = 1/3n(3) = 1/3n(n − 1)(n − 2) = 1/3n3 − n2 + 2/3n.
Just as each polynomial of degree k can be formed by adding multiples of the powers
n0 , n1 , . . . , nk , it can also be obtained by adding multiples of n(0) , n(1) , . . . , n(k) (see Exercise 8.3). Unfortunately, there is no easy way to compute the coefficients required to form
a given polynomial, even when it is a single power, from falling factorials. When we can
write a formula in terms of falling factorials, then finding antidifferences becomes easy, as
in the next example.
Example 8.1. Since we can write n2 as
n2 = n2 − n + n = n(n − 1) + n = n(2) + n(1) ,
we have
∆−1 (n2 ) = ∆−1 (n(2) + n(1) ) = 1/2n(3) + 1/2n(2)
= 1/3n(n − 1)(n − 2) + 1/2n(n − 1) = 1/3n3 − 1/2n2 + 1/6n.
In the calculus it is often useful to compute the derivative of a derivative, that is a second
derivative. In the same way we can define the second difference – the difference of the
difference – and higher order differences as well. We will denote these by powers
∆2 (f (x)) = ∆(∆(f (x))
∆3 (f (x)) = ∆(∆2 (f (x))),
and in general
∆k (f (x)) = ∆(∆k−1 (f (x))),
for k ≥ 1.
(Here k = 1 is included by adopting the convention that ∆0 = I, where I is the identity
operator with I(f (x)) = f (x) for all functions.)
48
§8. Falling Factorials
Thus, for example, we have
∆(n2 ) = 2n + 1
∆2 (n2 ) = ∆(∆(n2 )) = ∆(2n + 1) = 2,
and
∆3 (n2 ) = ∆(∆2 (n2 )) = ∆(2) = 0.
The higher differences of a falling factorial are easy to compute, since
∆(n(k) ) = kn(k−1) ,
∆2 (n(k) ) = ∆(kn(k−1) ) = k(k − 1)n(k−2) ,
∆3 (n(k) ) = ∆(k(k − 1)n(k−2) ) = k(k − 1)(k − 2)n(k−3) ,
and so forth. Note that for either nk or n(k) we would eventually reach a constant (for
n(k) it would be k(k − 1) . . . 1n(0) = k!) and that the (k + 1)st difference would be 0. The
same is true for polynomials in general. Each successive difference lowers the degree by
one until we reach a constant, and the next difference is 0.
If (an ) is a sequence in which the nth term is given by a polynomial, an = p(n), then
by continuing to take differences we will eventually reach a constant sequence, followed by
the sequence of all zeroes. If we have enough terms of the sequence, we can use them to
see how many steps it takes. This will tell us the degree of the polynomial as in the next
example.
Example 8.2. Suppose an = p(n) is a polynomial and that the sequence (an ) begins with
1, 7, 19, 37, 61, and 91. A convenient way to record the differences ∆(an ) and higher order
differences, ∆2 (an ), ∆3 (an ), etc., is to form a difference table where each term is placed
between and below the two terms from which it is computed. For this example we obtain
(an )
∆(an )
∆2 (an )
∆3 (an )
1
7
6
19
12
6
37
18
6
0
24
6
0
49
61
30
6
0
91
§8. Falling Factorials
Since the sequence of second differences, ∆2 (an ), is nonzero and constant, the preceding
sequence, ∆(an ), must be given by a 1st degree polynomial of the form An + B. In fact,
it must be of the form
∆(an ) = ∆−1 (6) = 6n + c
for some constant c. It follows that the original sequence is given by some quadratic
polynomial. We could use ∆−1 (n) = ∆−1 (n(1) ) = 1/2n(2) = 1/2n2 − 1/2n to compute
an = ∆−1 (6n + c) = 6( 1/2n2 − 1/2n) + cn + d,
and use the first terms of the sequence to find the constants c and d. It would be just as
easy, once we know that the polynomial has degree 2, to use three terms of the sequence
to compute all of the coefficients in p(n) = An2 + Bn + C, as in Example 7.2. We leave
the final steps to you (see Exercise 8.4).
Notice that for an exponential of the form r n , r 6= 1, we have ∆(r n ) = (r − 1)r n ,
∆2 (r n ) = ∆((r − 1)r n ) = (r − 1)2 r n , ∆3 (r n ) = (r − 1)3 r n , etc. Thus repeated differencing
will never produce 0 in this case.
Suppose we are given a sequence (an ). Can we be sure that there exists a sequence
(An ) which has (an ) as its sequence of differences? That is, must there exist (An ) for
which
an = ∆−1 (An )?
For the differential operator D the corresponding question is, given a function f (x), does
there exist a function F (x) with f (x) as its derivative? Do we have an antiderivative for
f (x)? Here the answer is ‘Yes,’ provided that f (x) is continuous. If it is, then the function
Z x
f (t)dt
F (x) =
a
satisfies D(F (x)) = F 0 (x) = f (x) for each choice of the constant a.
(Of course, we may not have a simple formula by which we can compute values of
Rx 2
2
F (x). We do not, for example, when f (x) = ex and F (x) = a et dt.)
For sequences antidifferences always exist. For any sequence (an ) there is an infinite
family of sequences (An ) with ∆(An ) = an . It is not hard to see how to find such sequences.
50
§8. Falling Factorials
Suppose we are given (an ), n ≥ 0. To construct An = ∆−1 (an ) we can begin with any
value A for A0 . The value for A1 must be such that ∆(A0 ) = a0 . This requires that
∆(A0 ) = A1 − A0 = a0 .
So we have
A1 = A 0 + a 0 = A + a 0 .
To obtain ∆(A1 ) = A2 − A1 = a1 , we take A2 to be
A2 = A 1 + a 1 = A + a 0 + a 1 .
In general we can insure that ∆(Ak ) = Ak+1 − Ak = ak by taking Ak+1 to be
Ak+1 = Ak + ak = A + a0 + a1 + · · · + ak = A +
X
aj .
0≤j≤k
This gives a sequence of antidifferences An = ∆−1 (an ). We have such a sequence for each
choice of A in the formula
An = A + a0 + a1 + · · · + an−1 = A +
X
aj .
0≤j≤n−1
To make use of this way of obtaining ∆−1 (an ) we would usually need some compact
formula for the sum of the aj , 0 ≤ j ≤ n−1. In cases where we already have An = ∆−1 (an )
we can reverse the process to find such formulae for finite sums, much as we can use
antiderivatives to compute definite integrals.
Suppose An = ∆−1 (an ). Then for each j ≥ 0 we have
aj = ∆(Aj ) = Aj+1 − Aj .
It follows that
X
aj = a0 + a1 + · · · + an = (A1 − A0 ) + (A2 − A1 ) + · · · + (An+1 − An ) = An+1 − A0 .
0≤j≤n
51
§8. Falling Factorials
All but two of the Aj appear twice and cancel out. For a sum which begins beyond a0 ,
say with ak , we get
X
aj = ak + ak+1 + · · · + an
k≤j≤n
= (Ak+1 − Ak ) + (Ak+2 − Ak+1 ) + · · · + (An+1 − An ) = An+1 − Ak .
This is the discrete analogue of that part of the Fundamental Theorem of Calculus which
states that if D(F (x)) = f (x), then
Z
b
f (x)dx = F (b) − F (a).
a
(Notice, however, that the sum of the aj ending with an is found using An+1 , rather than
An .)
P
Example 8.3. Consider the sum 0≤j≤n j 2 = 02 + 12 + 22 + · · · + n2 . Since this is the
P
sum 0≤j≤n aj for aj = j 2 , we have
X
j 2 = An+1 − A0 ,
0≤j≤n
for any choice of any antidifference sequence An = ∆−1 (n2 ). Since ∆−1 (n2 ) = ∆−1 (n(2) +
n(1) ) = 1/3n(3) + 1/2n(2) + c, we obtain
X
j 2 = An+1 − A0 = ( 1/3(n + 1)(3) + 1/2(n + 1)(2) ) − (0 + 0)
0≤j≤n
= 1/3(n + 1)(n)(n − 1) + 1/2(n + 1)(n) = 1/3n3 + 1/2n2 + 1/6n.
Exercises
8.1. Find the following differences.
i) ∆2 (n3 − 2n)
ii) ∆2 (n(4) )
iii) ∆k (n(k) ), k ≥ 0
8.2. Write n3 as a sum of multiples of n(1) , n(2) , and n(3) , and use this to find ∆−1 (n3 ).
8.3. Prove (by induction) that nk can be written as a sum of multiples of n(1) , n(2) , . . . , n(k) ,
for each k ≥ 0. (Do not expect to find explicit formulae for the nk .)
52
§8. Falling Factorials
8.4. Complete Example 8.2 by finding the polynomial p(n) for which an = p(n).
8.5. Find the
P following sums using antidifferences.
i)
j
0≤j≤n
P
ii)
j
10≤j≤n
P 3
iii)
j (see Exercise 8.2)
0≤j≤n
P j
3
iv)
0≤j≤n
8.6. Find ∆(4n ), ∆2 (4n ), and ∆3 (4n ).
8.7. Find a polynomial p(n) such that the sequence an = p(n) begins with
i) −1, 1, 7, 17, 31, 49
ii) 1, 2, 4, 8, 15
iii) 0, 0, 1, 3, 6, 10.
8.8. The rising factorials are defined by
n(k) = n(n + 1)(n + 2) . . . (n + k − 1),
k ≥ 1.
i) Find ∆(n(k) ).
ii) Find ∇(n(k) ), where ∇ denotes the backwards difference operator (see Exercise
6.7).
8.9. Show that ∆( 1/n(k) ) = (−k) 1/n(k+1) .
8.10. The result of Exercise 8.9 resembles the differentiation formula
D(x−k ) = D( 1/xk ) = −kx−k−1 = (−k)
1
xk+1
.
This suggests viewing 1/n(k) as a falling factorial with a negative subscript. This is
sometimes done, but since 1/n(k) is not defined for n = 0, the more common definition
of the falling factorials of negative order is
n(−k) =
1
1
1
=
=
(k)
(n + k)(k)
(n + 1)(n + 2) · · · (n + k)
(n + 1)
Show that with this definition we have
∆n(k) = kn(k−1)
for all integers k.
8.11. Find ∆−1 (n(−k) ) for k > 0. (Be careful with k = 1.)
53
for k > 0.
§8. Falling Factorials
8.12. Find ∆2 (n(−k) ) for k > 0.
8.13. Let f (x) be a polynomial of degree m. Show that if
g(x) =
m
X
∆k f (0)
k=0
k!
x(k) ,
then f (x) = g(x). (Here ∆k f (0) denotes the value of ∆k (f (x)) at x = 0. Consider
∆k g(0) for k = 0, 1, . . . , m + 1.)
This is known as Newton’s Interpolation Formula.
54
§9. Series
§9. Series
Consider the sum
Sn =
X
( 1/2)j = ( 1/2)0 + ( 1/2)1 + · · · + ( 1/2)n = 1 + 1/2 + 1/4 + · · · + 1/2n .
0≤j≤n
We can use antidifferences as in the last section to find a formula which gives the value of
Sn for all n, n ≥ 0. If An = ∆−1 (( 1/2)n ), then we have
X
Sn =
( 1/2)n = An+1 − A0 .
0≤j≤n
Remembering that ∆(r n ) = (r − 1)r n , for each r 6= 1 we have
∆−1 (r n ) =
1
rn ,
r−1
so in this case we can take An to be
An =
1/2
1
( 1/2)n ,
−1
which yields
Sn =
X
( 1/2)j =
0≤j≤n
1
( 1/2)n+1 − 1
1
1/2)n+1 −
1/2)0 =
(
(
.
1/2 − 1
1/2 − 1
1/2 − 1
These sums form a sequence (Sn ) and, since we can rewrite Sn as
Sn = 2(1 − ( 1/2)n+1 ),
it is not hard to see that this sequence converges. As n increases, the term ( 1/2)n+1 tends
to 0, so we have
lim Sn = 2.
n→∞
Now consider another sequence of such sums.
Sn =
X
3j = 3 0 + 3 1 + 3 2 + · · · + 3 n .
0≤j≤n
55
§9. Series
Using ∆−1 (3n ) =
1
3−1
· 3n , we obtain
Sn =
X
0≤j≤n
3j =
1
3n+1 − 1
1
· 3n+1 −
· 30 =
.
3−1
3−1
3−1
In this case Sn can also be written as
Sn = 3/2 · 3n − 1/2,
and it is clear that the sequence (Sn ) has no finite limit. This should come as no surprise
since these sums are being formed by adding higher and higher powers of 3.
We will call a sequence of sums of the form
X
Sn =
aj
0≤j≤n
an infinite series. The terms Sn will be called the partial sums of the series.
As the last two examples show, such a series may or may not have a finite limit. If it
does we say that the series is convergent and that it converges to this limit. Otherwise we
say that the series diverges and that it is divergent. Either way we will denote the series
by
X
aj
or
j≥0
X
aj .
0≤j≤∞
For a convergent series we interpret this expression as having as its value the number to
which the sequence of partial sums (Sn ) converges. In this case we can give a formal
P
meaning to the sum of all of the infinitely many members of the sequence (an ). If j≥0 aj
is convergent, by the sum of all aj , j ≥ 0 we mean the limit of the sequence of partial
sums.
We will say that the value of a divergent series is infinite and indicate this by writing
X
aj = ∞.
j≥0
With these interpretations our analysis of the last two examples yielded
X
( 1/2)j = 2,
j≥0
56
§9. Series
and
X
3j = ∞.
j≥0
These two examples represent the same type of series. Each is of the form
P
j≥0
aj where
in the sequence (an ) the ratio of any two consecutive terms is always the same. In the
first example an+1 /an = 1/2 for each n and in the second an+1 /an = 3. A series with this
property is called a geometric series.
If for all n ≥ 0 we have an+1 /an = r for some constant r, then a0 6= 0 and a1 = ra0 ,
a2 = ra1 = r 2 a0 , and so forth. It follows that an = a0 r n for each n. Each partial sum has
the form
Sn =
X
a0 r j = a0 (1 + r + r 2 + · · · + r n ) = a0
0≤j≤n
X
rj .
0≤j≤n
In the special case where r = 1 we have an = a0 for each n ≥ 0, and the partial sums are
Sn = a0 (1 + 1 + · · · + 1) = (n + 1)a0 .
In this case the series diverges.
For r = −1 we have an = (−1)n a0 . The partial sums begin with S0 = a0 , S1 =
a0 − a0 = 0, S2 = a0 − a0 + a0 = a0 , and S4 = a0 − a0 + a0 − a0 = 0. Since this sequence
continues to alternate between a0 and 0, there is no limit.
For each r 6= 1, just as in the examples with r = 1/2 and r = 3, we can use the
antiderivative ∆−1 (r n ) =
P
j
j≥0 ar are of the form
1
r−1
r n to show that the partial sums of the geometric series
Sn = a
r n+1 − 1
.
r−1
(This can also be seen directly by noting that almost all of the terms of rSn − Sn cancel
giving rSn − Sn = ar n+1 − a · 1.)
If |r| < 1, then r n+1 tend to 0 as n increases. In this case
lim Sn =
n→∞
a
.
1−r
If |r| > 1, then |r n+1 | grows without bound as n increases.
57
§9. Series
Together, these facts show that the geometric series
P
j≥0
ar j converges to a/1−r if
|r| < 1 and diverges if |r| ≥ 1. There is another way to see that the geometric series
P
j
j≥0 ar must diverge if |r| ≥ 1. The difference between consecutive partial sums is
|Sn+1 − Sn | = |a0 r n+1 | = |a0 ||r|n+1 .
For |r| ≥ 1 these differences are always at least |a0 | (and for |r| > 1 they are increasing).
If the sequence (Sn ) has a limit, the difference between successive terms must tend to 0.
P
Suppose
Then the sequence of partial sums has a limit,
j≥0 aj converges.
limn→∞ (Sn ) = S. Then for each > 0 there is an integer N such that |Sn − S| < for all n ≥ N . This gives
|Sn+1 − Sn | = |(Sn+1 − S) − (Sn − S)| ≤ |Sn+1 − S| + |Sn − S| < 2.
Since |Sn+1 − Sn | = |an+1 |, this means that we must have limn→∞ an = 0. (We
could begin with the positive number /2 and show that there is an integer N such that
|an+1 − 0| < for all n ≥ N .) This proves the following:
P
If j≥0 aj converges, then limn→0 an = 0.
Warning: This actually gives us a test for divergence, not for convergence. It says that
if limn→∞ an does not exist, or if it exists but is not 0, then the series must diverge. It
does not say that we have convergence for every sequence with limn→∞ an = 0. Our next
example illustrates this.
Example 9.1. The series
X
1/n
= 1 + 1/2 + 1/3 + · · · + 1/n + · · ·
j≥1
is known as the harmonic series. The first few partial sums are
H1 = 1
H2 = 1 + 1/2 = 3/2 = 1.5,
H3 = 1 + 1/2 + 1/3 =
and
58
11/6
≈ 1.83,
§9. Series
H4 = 1 + 1/2 + 1/3 + 1/4 =
25/12
≈ 2.08.
The sequence of partial sums (Hn ) is increasing. To decide if the harmonic series converges,
we must determine whether this sequence has a finite limit.
Notice that for each n ≥ 1 we can group the terms of the sum H2n as follows:
H2n = 1 + ( 1/2) + ( 1/3 + 1/4) + ( 1/5 + 1/6 + 1/7 + 1/8) + · · · + ( 1/2n−1 +1 + · · · + 1/2n ).
The sum of the terms inside any one pair of parentheses is at least 1/2 since it consists
of some 2k−1 terms, each at least 1/2k . For example, 1/3 + 1/4 ≥ 1/4 + 1/4 = 1/2 and
1/5
+ 1/6 + 1/7 + 1/8 ≥ 4( 1/8) = 1/2. Since there are n such groups for H2n , we have
H2n ≥ 1 + n( 1/2).
Although the sequence (Hn ) grows very slowly (about 215 terms are actually needed to
exceed 10), its terms will eventually exceed any fixed value, so there can be no limit. It
follows that the harmonic series diverges.
Note: We have limn→∞ an = 0 both for an = 1/n and for an = r n when |r| < 1.
P
But the series j≥0 r j converges for |r| < 1, while the harmonic series diverges. What we
P
know so far is that if limn→∞ an 6= 0, then the series j≥0 aj diverges. If limn→∞ an = 0,
we may or may not have convergence for the series, and we will have to look further.
P
The partial sums of the harmonic series, Hn = 1≤j≤n 1/j , are themselves called the
harmonic numbers. For an integer of the form N = 2n , where n = log2 N , the inequality
used to show that (Hn ) has no finite limit can be rewritten as
HN ≥ 1 + 1/2 log2 n.
This suggests a connection between these partial sums and logarithmic functions. Actually
we made use of such a connection earlier to show that for sequences we have log 2 n = o(n).
The key ideas were that, first, logb n = ln n/ ln b, where ln n denotes the natural logarithm
with base e, and second, that estimates of ln n can be obtained by approximating the
Rx
integral 1 1/tdt which defines ln x. The approximation of ln n was obtained by enclosing
the area beneath the graph of f (t) = 1/t from t = 1 to t = n by rectangles and produced
ln n < 1 + 1/2 + 1/3 + · · · + 1/n−1 = Hn−1 .
59
§9. Series
This, together with the fact that f (x) = ln x increases without bound as x increases, gives
another proof that the harmonic series diverges.
We can use the connection between partial sums of a series and approximations to
integrals to investigate the convergence or divergence of the series in other cases as well.
We do this in the next example.
Example 9.2. Consider the series
P
j≥1
1/j 2 .
Since we have limn→∞ 1/n2 = 0, there is a
chance that this series converges, but remember that this is not enough. The partial sums
of the series have the form
An =
X
1/j 2
= 1 + 1/22 + 1/32 + · · · + 1/n2 .
1≤j≤n
Since An is the sum of the values of the function f (t) = 1/t2 at t = 1, 2, 3, . . ., and n, it is
related to the integral of this function from t = 1 to t = n. In fact we can approximate
this integral both from above and below by sums of this form. If we enclose the area under
the graph of f from t = 1 to t = n by rectangles of width one we obtain
1
2
2
1
1
2
n−1
3
n
which gives
Z
n
1/t2 dt
< 1 + 1/22 + 1/32 + · · · + 1/(n−1)2 = An−1 .
1
By drawing rectangles below the graph we get
60
§9. Series
1
and therefore
Z
Since we have
Z
we have shown that
n−1
3
n
n
1/t2 dt
> 1/22 + 1/32 + · · · + 1/n2 = An − 1.
1
n
1/t2 dt
1
2
t=n
= 1/(−1) t−1 t=1 = [− 1/n − (−1)] = 1 − 1/n,
An >
Z
n+1
1/t2
dt = 1 −
1
and
An < 1 +
Z
1
,
n+1
n
1/t2
dt = 2 − 1/n.
1
The first inequality tells us that the terms of the sequence of partial sums are larger than
those of the sequence (1 − 1/n+1) which has limit 1. This could happen if (An ) itself has a
limit (which would then have to be at least 1), but it could also be the case if (An ) has no
limit at all. For example, we have n > 1 − 1/n+1, and even ln n > 1 − 1/n+1, for all n ≥ 2,
but neither (n) nor (ln n) has a limit. We cannot draw any conclusions from the fact that
(An ) is term by term larger than a convergent sequence.
The second inequality tells us that the terms of (An ) stay below those of the convergence sequence (2 − 1/n). For all n ≥ 1 we have An ≤ 2 − 1/n ≤ 2. The sequence (An ) of
partial sums is both increasing (since An = An−1 + 1/n2 ) and bounded, so it must converge.
P
We have shown that the series j≥1 1/j 2 converges, and furthermore that
X 1
< 2.
j2
j≥1
61
§9. Series
Note. This reasoning can be used to establish a general comparison test for two series
with nonnegative terms. For such series we have
P
P
i) If aj ≤ bj for each j ≥ 0 and j≥0 bj converges, then j≥0 aj converges.
P
P
ii) If aj ≤ bj for each j ≥ 0 and j≥0 aj diverges, then j≥0 bj diverges.
This test can also be used if we have aj ≤ bj for all j ≥ N for N 6= 0. But we must
know that aj and bj are (eventually) nonnegative so that the sequences of partial sums are
increasing. For details see Exercises 9.5 and 9.6.
Notice also that we can draw no conclusions about
P
j≥0
aj from the information that
the aj are larger than the terms of a series which converges or smaller than those of a
series which diverges.
The reasoning of Examples 9.1 and 9.2 lead to another general principle known as
the integral test. Suppose f (t) is a positive decreasing function. Just as for the function
f (t) = 1/t2 in Example 9.2, we have
Z
n
f (t)dt ≤ f (1) + f (2) + · · · + f (n − 1),
1
and
Z
n
f (t)dt ≥ f (2) + f (3) + · · · + f (n).
1
Rn
P
f
(t)dt
diverges,
then
the
sequence
of
partial
sums
of
the
series
j≥1 f (j)
1
Rn
also diverges. If the sequence 1 f (t)dt converges, then these partial sums also converge,
P
and j≥1 f (j) is a convergent series.
Rn
R∞
Since the sequence 1 f (t)dt converges if the infinite integral 1 f (t)dt converges, and
R∞
is unbounded if 1 f (t)dt diverges, we can state this as follows:
R∞
P
If f is a positive decreasing function, then the series j≥1 f (j) converges if 1 f (t)dt
If the sequence
converges, and diverges if this integral diverges.
This test can be used to show that the series
P
j≥1
p > 1, and diverges when 0 < p ≤ 1 (see Exercise 9.7).
Exercises
9.1. Find the following partial sums.
62
aj , aj = 1/j p , converges for each
§9. Series
i)
P
1/4j
0≤j≤n
ii)
P
(2j + 1)
0≤j≤n
iii)
P
(2j − j)
0≤j≤n
9.2. Determine
P √whether the series converges or diverges.
1/ j
i)
j≥1
P
ii)
( 1/3)j
j≥0
P
1/j 2 +1
iii)
j≥0
P
1/j ln j
iv)
j≥2
P
1/log j
v)
2
j≥2
P
P
9.3. Give a convergent series j≥0 aj and divergent series j≥0 bj , both with nonnegative
terms, such that aj ≤ bj for all j ≥ 0.
9.4. Find a sequence
(an ) with an = o( 1/n) for which the series
P
for which j≥1 aj converges.
P
j≥1
aj diverges and one
P
P
9.5. Suppose that j≥0 aj andP j≥0 bj are series of nonnegative
terms for which aj ≤ bj
P
for all j ≥ 0. Show that if j≥0 bj is convergent with j≥0 bj = B,Pthen the sequence
of partial sums of this series is bounded, and use this to show that j≥0 aj converges.
P
P
9.6. Suppose that j≥0 aj and Pj≥0 bj are series of nonnegative terms for which aj ≤ bj
for all j ≥ 0. Show that if j≥0 aj is P
divergent, then its sequence of partial sums is
unbounded, and use this to show that j≥0 bj diverges.
Rn
P
9.7. Compute the integral 1 1/tp dt for p 6= 1. Use this to show that the series j≥1 1/j p
converges for each p > 1 and diverges if 0 < p ≤ 1.
9.8. Determine
P j−1whether the series converges or diverges.
i)
2j
j≥1
ii)
P
j≥1
iii)
P
j≥0
| sin j|
2j
1
j!
9.9. If the series
P
j≥0
aj converges, must the series
9.10. If the series
P
j≥0
|aj | converges, must the series
63
P
j≥0
P
|aj | also converge?
j≥0
aj also converge?
§9. Series
P
1
9.11. Show that the series
j≥1 j(j+1) converges and find its value. (Use the fact that
1
1
= 1j − j+1
to find a formula for the partial sums of this series.)
j(j+1)
9.12. Show that ln(n!) ≥
9.13. Does the series
P
Rn
1
ln tdt, and use this to show that n! ≥ e( n/e)n .
j≥1 (1
− 1/j )j converge? Explain.
P
9.14. Suppose j≥0
Paj is a series with non-negative terms. Show that if this series converges,
then so does j≥0 a2j .
9.15. Show by an example that the conclusion of Exercise 9.14 may be false if we do not
require the terms of the series to be non-negative.
9.16. For what values of n is Hn > 1000?
64
§10. Evaluating Series
§10. Evaluating Series
We now have several methods for trying to determine whether a series converges or
diverges. If the series does converge, these same ideas can sometimes be used to obtain
information about its value.
P
Example 10.1. Consider the series j≥0 j 31+1 . The terms of this series resemble those of
P
P
P
1 3
1 p
1 3
j≥1 /j , which converges since it is of the form
j≥1 /j , with p > 1. Since
j≥1 /j
converges and 1/j 3 +1 < 1/j 3 for each j ≥ 1, we can use the comparison test to show that
the original series converges. Notice that although this series has one additional term for
P
j = 0, we can still apply this method. The partial sums of j≥1 1/j 3 +1, starting at j = 1,
P
converge by comparison with those of j≥1 1/j 3 . Adding the term for j = 0 increases each
of these partial sums by 1, so the limit still exists.
Now we could combine the inequality
X
0≤j≤n
with estimates of
P
1≤j≤n
1/j 3
X 1
1
1
<
+
j3 + 1
1
j3
1≤j≤n
derived from integrals of f (t) = 1/t3 to obtain upper bounds
for the value of our series. (We could also use integrals of f (t) = 1/t3 +1 directly, but the
integration is much more complicated.)
This way we obtain
X
1≤j≤n
1
1
< 3+
3
j
1
Z
n
1
n
1
dt = 1 + − 1/2t−2 1 = 1 +
3
t
1/2
1
− 2 < 3/2,
2n
and therefore for each n
X
0≤j≤n
j3
1
< 1 + 3/2 = 2.5.
+1
Since all partial sums are bounded by 2.5, so is their limit, and we have
X
j≥0
Since the terms of
P
j≥0
1/j 3 +1
j3
1
≤ 2.5.
+1
are non-negative, its partial sums are increasing and each
of them therefore gives a lower bound for the limit. For example, we have
X
j≥0
X
1
1
1
3
>
=
1
+
=
= 1.5,
j3 + 1
j3 + 1
13 + 1
2
0≤j≤1
65
§10. Evaluating Series
X
j≥0
X
3
1
29
1
1
>
= + 3
=
≈ 1.611,
3
3
j +1
j +1
2 2 +1
18
0≤j≤2
and
X
j≥0
X
1
1
29
1
415
>
=
+ 3
=
≈ 1.647.
3
3
j +1
j +1
18 3 + 1
252
0≤j≤3
Taking more terms would give better bounds, but these partial sums do not grow very
quickly. This suggests that we might be able to improve the upper bound.
P
P
Since the terms of j≥0 1/j 3 +1 and those of j≥0 1/j 3 differ most when j is small,
and the largest errors in the integral approximations also occur for small j in this case,
we can do better by using partial sums and integrals which begin beyond j = 1. Starting
with j = 2 gives
X
X 1
1
1
1/2 +
3/2 +
=
1
+
<
j3 + 1
j3 + 1
j3
2≤j≤n
2≤j≤n
0≤j≤n
Z n
1
1
dt = 3/2 + 1/8 + 1/8 − 2 < 1.75,
< 3/2 + 1/23 +
3
2n
2 t
X
while starting with j = 3 would give
X
X 1
1
1
1/2 + 1/9 +
29/18 +
=
1
+
<
j3 + 1
j3 + 1
j3
0≤j≤n
3≤j≤n
3≤j≤n
Z n
1
1
< 29/18 + 1/33 +
dt = 89/54 + 1/18 − 2 <
3
2n
3 t
X
92/54
≈ 1.704,
for all n.
A similar computation using
P
4≤j≤n
X
j≥0
j3
1/j 3
would yield the upper bound
1
≤ 1.681,
+1
while a partial sum with 10 terms gives
X
j≥0
X
1
1
>
≈ 1.676.
3
3
j +1
j +1
0≤j≤9
This tells us that the sum of the first 10 terms already gives us a good approximation to
the sum of the entire series. We can also use this approach to obtain a bound on the error
which results when we do use a partial sum to estimate the value of the series.
66
§10. Evaluating Series
For each positive integer N and all n ≥ N we have
X
X
X
X 1
X
1
1
1
1
=
+
<
+
j3 + 1
j3 + 1
j3 + 1
j3 + 1
j3
0≤j≤N −1
N ≤j≤n
0≤j≤N −1
N ≤j≤n
0≤j≤n
Z n
X
1
1
1
<
+ 3+
dt
3
3
j +1 N
N t
0≤j≤N −1
<
1
1
1
+
+
.
j3 + 1 N 3
2N 2
X
0≤j≤N −1
Since this is true for all sufficiently large partial sums, we have that


X
X
X
1
1
1
1
1 
<
≤
.
+ 3+
3
3
3
j +1
j +1
j +1
N
2N 2
0≤j≤N −1
j≥0
0≤j≤N −1
Using the first 10 terms (0 ≤ j ≤ 9) yield an error of at most
1
1
+
= 0.021,
3
10
2 · 102
while using the first 100 terms would reduce the error to at most
1
1
+
≈ 0.0005.
6
10
2 · 104
Since we could compute the sum of a large number of terms of this series quickly using
a computer or programmable calculator, it would not be hard to obtain estimates of the
value of the series with almost any desired degree of accuracy.
Since the value of a convergent series is the limit of its sequence of partial sums,
finding the exact value to which it converges usually requires a suitable formula for those
partial sums. Even for relatively simply series this is often difficult, if not impossible. We
can obtain such a formula, however, when we can recognize the terms of the series as a
sequence of differences. We have seen that if an = ∆(An ), or equivalently, An = ∆−1 (an ),
then
X
aj =
j≥0
For example, we have used ∆
X
∆(Aj ) = An+1 − A0 .
0≤j≤n
−1
n
(r ) =
1
n
r−1 r
for r 6= 1 to show that for a geometric series
we have
X
j≥0
j
ar = a
1
1
r n+1 −
r0
r−1
r−1
and therefore that such a series converges to a
1
1−r
=a
1 − r n+1
,
1−r
for |r| < 1.
Our next example pushes the use of antidifferences a bit further.
67
§10. Evaluating Series
Example 10.2. Consider the series
P
j≥0
j/2j .
Since n = o(2n ), the terms of this series
tend to 0, so it may converge. But, as usual, note that this is not enough. (Always
remember the divergent harmonic series.)
It is easy to see that j/2j ≥ 1/2j for all j ≥ 1, but this comparison doesn’t help. We
cannot conclude anything about convergence from the fact that the terms are as large or
larger than those of some convergent series. Nonetheless, since 2j grows rapidly, we may
expect to be able to show that the terms j/2j are eventually smaller than those of some
convergent series. In fact, it is not hard to show that j/2j ≤ 1/j 2 for sufficiently large j.
This is equivalent to showing that
j 3/2j
≤ 1. Since j 3 = o(2j ), and 1 is positive, we have
j 3/2j
< 1 when j is large enough. (We could also show that j 3 ≤ 2j for all j ≥ 8 by
P
P
induction.) Since j≥0 1/j 2 converges, so does j≥0 j/2j by the comparison test.
For this example we can actually find a formula for the partial sums, and use it to
find the exact value to which the series converges. The first step is to rewrite the terms of
the series as products.
We have
X
j≥0
j/2j
=
X
j( 1/2)j .
j≥0
Our object now is to find the antidifference ∆−1 (n( 1/2)n ). That is, we seek a sequence
(An ) with
∆(An ) = n( 1/2)n .
The new feature is that we are looking for the antidifference of a product. This corresponds
in the calculus to seeking the antiderivative of a product. The difficulty in integrating a
product is the result of the way we find the derivative of a product. The derivative of
a product is not simply the product of the derivatives of its factors. As we will see, the
difference of a product is not the product of differences either. The correct rule for finding
the derivative of a product does lead to a result about integrals of products which is one of
the most important facts about integration. This is the idea of integration by parts. Using
our operator notation for the differential operator, D(f ) = f 0 , the derivative of a product
is given by
D(f · g) = f D(g) + gD(f ).
68
§10. Evaluating Series
Taking integrals over both sides yields
Z
Z
Z
f · g = D(f · g) = f D(g) + gD(f ),
and therefore
Z
f D(g) = f g −
Z
For definite integrals this becomes
Z b
Z
b
f D(g) = [f g]a −
a
gD(f ).
b
gD(f ).
a
R
The main point of this formula is that it may allow us to find the integral f D(g) by
R
computing the (hopefully) more simple integral gD(f ). To use it we must be able to
R
recognize our original integral as being of the form f D(g) for some f and g.
To see how similar reasoning might be used for antidifferences, we must start by
figuring out how to find the difference of a product. We need a formula which corresponds
to the product rule for derivatives.
Suppose we apply the difference operator to the sequence obtained by multiplying
corresponding terms of the sequences (An ) and (Bn ). We get
∆(An Bn ) = An+1 Bn+1 − An Bn .
We want to express this in terms of the differences ∆(An ) = An+1 − An and ∆(Bn ) =
Bn+1 − Bn . We introduce differences we need by subtracting a term and adding it back
in.
∆(An Bn ) = An+1 Bn+1 − An Bn+1 + An Bn+1 − An Bn .
This can now be written as
∆(An Bn ) = An (Bn+1 − Bn ) + Bn+1 (An+1 − An ),
or
∆(An Bn ) = An ∆(Bn ) + Bn+1 ∆(An ).
Note the similarity to D(f g) = f D(g)+gD(f ). But also note that the second term involves
Bn+1 rather than Bn . This is an important feature of this product rule for differences,
69
§10. Evaluating Series
but it is easy to overlook. (In the derivation of the product rule for derivatives there is a
corresponding term, g(x + ∆x), which leads to g(x) in the limit as ∆x → 0.)
Taking sums on both sides yields
X
X
X
∆(Aj Bj ) =
Aj ∆(Bj ) +
Bj+1 ∆(Aj ),
0≤j≤n
0≤j≤n
0≤j≤n
and therefore
X
0≤j≤n
Aj ∆(Bj ) =
X
∆(Aj Bj ) −
0≤j≤n
X
Bj+1 ∆(Aj ).
0≤j≤n
Now since the middle term is a sum of differences we get
X
X
Aj ∆(Bj ) = (An+1 Bn+1 − A0 B0 ) −
Bj+1 ∆(Aj ).
0≤j≤n
0≤j≤n
This is the rule for summation by parts. (Again note that the last term involves B j+1
rather than Bj .)
We can try to use summation by parts whenever we are dealing with a sum whose
terms are products of the form Aj ∆(Bj ) for some (An ) and (Bn ). We must be able
to spot the differences ∆(Bj ). And of course, we need to be able to find the new sum
P
0≤j≤n Bj+1 ∆(Aj ).
To apply summation by parts to our example, we need to find sequences (An ) and
(Bn ) such that
An ∆(Bn ) = n( 1/2)n .
There are two obvious ways to choose (An ) and (Bn ).
1. Set An = n and ∆(Bn ) = ( 1/2)n , or
2. set An = ( 1/2)n and ∆(Bn ) = n.
In the first case we have ∆(An ) = 1 and Bn = ∆−1 ( 1/2)n =
would lead to a new sum with terms of the form
1
1 n
1/2−1 ( /2)
= −2( 1/2)n . This
Bj+1 ∆(Aj ) = −2( 1/2)j+1 · 1 = −( 1/2)j
The second choice produces ∆(An ) = −2( 1/2)n and Bn = ∆−1 (n) = ∆−1 (n(1) ) = 1/2n2 −
1/2n.
The terms of the new sum would be of the form
Bj+1 ∆(Aj ) = ( 1/2(j + 1)2 − 1/2(j + 1))(−2( 1/2)j ).
70
§10. Evaluating Series
In the second case we get terms that are more complicated than those of the original series.
So we use the first. Substituting into
X
Aj ∆(Bj ) = (An+1 Bn+1 − A0 B0 ) −
0≤j≤n
X
Bj+1 ∆(Aj )
0≤j≤n
gives
X
j( 1/2)j = ((n + 1)(−2)( 1/2)n+1 − 0(−2)( 1/2)0 ) −
0≤j≤n
X
(−( 1/2)j ) · 1
0≤j≤n
X
n
= −(n + 1)( 1/2) +
( 1/2)
j
0≤j≤n
In place of the original sum, we have a partial sum of a geometric series. Using
1−rn+1
1−r
for that sum we obtain
X
P
0≤j≤n
rj =
j( 1/2)j = −(n + 1)( 1/2)n + 2(1 − ( 1/2)n+1 ) = 2 − (n + 2)( 1/2)n .
0≤j≤n
Since (n + 2) = o(2n ), we have limn→∞ (n + 2)( 1/2)n = 0. It follows that
X
j/2j
= 2.
j≥0
For any r 6= 1 summation by parts with An = n and ∆(Bn ) = r n leads to
X
jr j = −(n + 1)
X
r
1
r n+1 +
rj
1−r
1−r
0≤j≤n
0≤j≤n
and hence for |r| < 1 to
X
j≥0
jr j = 0 +
r
1
r
·
=
.
1−r 1−r
(1 − r)2
Nearly all of the sequences we have seen so far have had non-negative terms. In fact,
the convergence criteria known as the comparison test and the integral test only apply to
series of non-negative terms. Series with both positive and negative terms are typically
more complicated. A special case which is often more manageable is that of an alternating
71
§10. Evaluating Series
series in which the terms strictly alternate between positive and negative values. A good
example is the alternating harmonic series.
X
(−1)j+1 ( 1/j ) = 1 − 1/2 + 1/3 − 1/4 + − · · · .
j≥1
If a series converges, then its terms must tend to 0, but this alone does not insure conP
vergence. For an alternating series
aj the condition limj→∞ aj = 0, together with one
other property, does guarantee convergence. The full statement is the following:
An alternating series
(|aj |) is decreasing
P
j≥0
aj for which limj→∞ aj = 0 and such that the sequence
|a0 | ≥ |a1 | ≥ |a2 | ≥ |a3 | ≥ · · · ≥ |aj | ≥ · · ·
is convergent. Thus, for example, the alternating harmonic series converges, while the
ordinary harmonic series diverges.
To see why this is so, consider an alternating series
P
j≥0 (−1)
j
aj with a0 > 0. (Alter-
nating series which begin with a negative term can be treated in the same way.) Suppose
that limj→∞ aj = 0 and that (aj ) is a decreasing sequence.
First note that half of the partial sums can be written as
S2n+1 =
X
(−1)j aj = (a0 − a1 ) + (a2 − a3 ) + · · · + (a2n − a2n+1 ).
0≤j≤2n+1
Since (aj ) is decreasing, each a2j − a2j+1 is non-negative, so the S2n+1 form an increasing
sequence. But S2n+1 can also be written as
S2n+1 = a0 + (−a1 + a2 ) + (−a3 + a4 ) + · · · + (−a2n−1 + a2n ) − a2n+1 .
Since each (−a2j−1 +a2j ) is at most zero, this shows that S2n+1 ≤ a0 for each n. Therefore
the S2n+1 form an increasing sequence which is bounded above by a0 and must have a
limit
lim S2n+1 = S.
n→∞
72
§10. Evaluating Series
The remaining partial sums can be written as
S2n = a0 + (−a1 + a2 ) + (−a3 + a4 ) + · · · + (−a2n−1 + a2n )
and form a decreasing sequence. Since S2n = S2n+1 − a2n+1 we have
lim S2n = lim S2n+1 − lim a2n+1 = S − 0 = S.
n→∞
n→∞
n→∞
Since the two sequences of partial sums have the same limit, the series converges to this
value.
If we plot the partial sums we obtain the pattern
S
S1 = a 0 − a 1
S3
S5
Sn+1
Sn
S4
S2
S0 = a 0
← |an+1 | →
The partial sums oscillate and close in on the limit S from both sides. For each n we have
|S − Sn | ≤ |an+1 |. The absolute value of the remainder of the series beyond Sn is at most
|an+1 |.
We can get a better estimate of S by taking the midpoint 1/2(Sn +Sn+1 ) of the interval
between Sn and Sn+1 . The difference between S and 1/2(Sn + Sn+1 ) is at most the radius
of that interval, 1/2|Sn+1 − Sn | = 1/2|an+1 |. Thus if |an+1 | ≤ , using 1/2(Sn + Sn+1 ) results
in an error of at most /2.
With the alternating harmonic series, for example, we have
X
S9 =
(−1)j+1 ( 1/j ) ≈ 0.74563
1≤j≤9
S10 =
X
(−1)j+1 ( 1/j ) ≈ 0.64563,
1≤j≤10
since |a10 | = | − 1/10| = 0.1, the value of the series differs from 1/2(S9 + S10 ) ≈ 0.69563 by
at most 0.05. Actually the error here is much less as we will soon see:
X
(−1)j−1 ( 1/j ) = ln 2 ≈ 0.69315.
j≥1
73
§10. Evaluating Series
Exercises
10.1. Find an approximation to
0.1.
P
1
j≥0 j 2 +1
which is accurate to within an error of at most
10.2. Determine
whether the series converges.
P
i)
(−1)j ln1j
j≥2
P
j
ii)
(−1)j 3j+1
j≥0
P
iii)
(−1)j lnj j
j≥1
P
iv)
(−1)j 2jj
j≥0
10.3. Evaluate
P kthe series.
i)
3k
k≥0
P 2
ii)
j ( 1/2)j
j≥1
P
iii)
j(2) · j
j≥1
10.4. Find P
all real numbers x for which the series converges.
i)
xj
j≥0
P
ii)
(−1)j xj , x > 0
j≥0
P
iii)
(−1)j xj , x < 0
j≥0
P
P
10.5. a) If j>0 aj converges, must j>0 a2j converge?
P
P
b) If j≥0 a2j converges, must j≥0 aj converge?
10.6. Evaluate
using antidifferences.
P the sums
2
i)
(k − k)
P0≤k≤n 2
ii)
0≤k≤n (k + k)
10.7. FindP
an approximation to the series with an error of at most 0.01.
j 1
i)
j≥1 (−1) 2j+1
P
j 1
ii)
j≥1 (−1) ( /j!)
10.8. The product rule for differences appears to lack symmetry.
∆(an bn ) = an ∆(bn ) + bn+1 ∆(an )
∆(bn an ) = bn ∆(an ) + an+1 ∆(bn )
74
§10. Evaluating Series
Are both of these equations correct? How is this possible?
10.9. Evaluate
sums (see Exercise 8.10).
P the following
1
i)
P1≤k≤n (k+1)(k+2)
1
ii)
1≤k≤n k(k+1)
10.10. Evaluate the series
P
k≥1
1
(k+1)(k+2)(k+3)
10.11. Let
bk =
(see Exercise 8.10).
1
1
1
− k(−3) = 3 −
3
k
k
(k + 1)(k + 2)(k + 3)
i) FindP
b1 , b2 , b3 and
P b4 .
P
P
1
ii) Use k≥1 k3 = k≥1 k(−3) + k≥1 bk and Exercise 10.10 to estimate k≥1 1/k3 .
10.12. Repeat Exercise 10.11 using
10.13. Find a formula for
P
1≤j≤n
1
(k+1)(3)
=
1
(k+1)(k)(k−1)
in place of k(−3) .
jHj , where Hj denotes the j th harmonic number.
75
Solutions 1.
1.1. If limn→∞ an = L1 and limn→∞ an = L2 , show that if L1 6= L2 then there is an > 0
such that |an − L1 | < and |an − L2 | < cannot both be true for the same n. E.g.
= 1/10|L1 − L2 |. Why does this settle the question?
1.2. Since
1
24
<
1
,
10
any N ≥ 4 will do.
1.3. i) Show that if limn→∞ an = L, then limn→∞ |an | = |L| by using the fact that
||an | − |L|| ≤ |an − L| (consider the distance between |an | and |L| on the real line).
ii) For example consider the sequence 0, 1, 0, 1, 0, 1, . . ..
1.4. i) Converges to L = 0.
ii) Converges to L = 1.
iii) Diverges.
iv) Converges to 2. (Show that an = 1 +
v) Converges to L = 1/2.
vi) Diverges.
2n −1
.)
2n
1.5. Show that if limn→∞ an = L, then limn→∞ (an + 1) = L + 1 by comparing |an − L|
and |(an + 1) − (L + 1)|.
1.6. Show that if limn→∞ an = L, then limn→∞ 2an = 2L by comparing |an − L| and
|2an − 2L| = |2||an − L| = 2|an − L|.
1
Solutions 2.
2.1. i) Converges to L = 2.
ii) Converges to L = 1.
iii) Converges to L = 0. 1
1
= n(n+1)
iv) Converges to L = 0. n1 − n+1
v) Converges to L = 0.
vi) Converges to L = 1 · 2 = 2. (See Theorem 1 (iii).)
2.2. i) For example, 0, 1, 0, 1, . . . and 1, 0, 1, 0, . . .
ii) For example, an = 2n , bn = 3n .
2.3. Suppose that limn→∞ an = L and limn→∞ bn = J . It is enough show that L < J is
impossible. Show that if J − L > 0, then eventually we have both
|bn − J | < 1/2(J − L)
and
|an − L| < 1/2(J − L),
so bn > J − 1/2(J − L) = 1/2(J + L) and an < L + 1/2(J − L) = 1/2(J + L).
2.4. Show that for each > 0 eventually we have both an > a − and bn < 0 + = , so
a
an /bn > a−
= − 1. Show that for each B there is an with a/ > B + 1.
2.5. i) Let bn = ln an , where ln denotes the natural logarithm. Then bn = − 1/n and
limn→∞ bn = 0, so limn→∞ an = e0 = 1. Or use Theorem 3 with f (x) = e−1/x .
ii) limn→∞ an = ( 1/3)1/2 . (See Example 3.)
2
Solutions 3.
3.1. i) The sequence is unbounded since for any B > 0 we have ( 3/2)n > B for all n with
n log2 ( 3/2) > log2 (B). (log2 ( 3/2) = log2 3 − log2 2 ≈ 1.6 − 1 = 0.6.)
ii) an ≤ 1 for all n. (Show that this is so by induction.)
iii) The sequence is unbounded. Show, for example, that for n ≥ 3 we have an ≥
( 3/2)n and use part (i). (If an−1 ≥ ( 3/2)n−1 and an ≥ ( 3/2)n , then an+1 ≥
( 3/2)n + ( 3/2)n−1 = 5/2( 3/2)n−1 > 9/4( 3/2)n−1 = ( 3/2)n+1 .)
1
≥ nk n1k for 0 ≤ k ≤ n, but not without a
3.2. It can be shown directly that n+1
k
(n+1)k
fair amount of computation. Alternatively, one could use induction on the value of k.
There is equality for k = 0 and k = 1. For the inductive step use the facts that
1
1
n+1
n+1
n−k+1
1
=
·
,
k+1
k + 1 (n + 1)
k
k+1
n+1
(n + 1)k
and show that
as well.)
n−k+1
k+1
·
1
n+1
and
n−k 1 n 1
n
1
=
·
,
k + 1 n k nk
k + 1 nk+1
≥
n−k
k+1
· n1 . ((1 +
1
n+1
n+1 )
has an additional positive term
1
by induction on k for k ≥ 2 using the second equation in
3.3. Show that nk · n1k < 2k−2
the solution of Exercise 3.2. This show that
(1 + 1/n)n < 1 + 1 +
1
1
1
+ 2 + · · · + n−2 < 3.
2 2
2
3.4. i) limn→∞ an = e2 . (See the note.)
ii) limn→∞ an = e−1 = 1/e.
2
iii) limn→∞ an = 1. (1 ≤ (1 + 1/n2 )n = ((1 + 1/n2 )n )1/n < e1/n .)
3.5. i) For a sequence of positive terms limn→∞ an+1 /an = L > 1 implies that there exists
an integer N such that for all n ≥ N
an+1 > an
L+1
,
2
k
, and therefore that the sequence
where L+1
> 1. It follows that aN +k > aN L+1
2
2
diverges.
ii) an = 1 − ( 1/2)n .
1
iii) an = 1 + 1/2 + 1/3 + · · · + 1/n, n ≥ 1. limn→∞ aan+1
= 1 since 1 < aan+1
≤ 1 + n+1
.
n
n
See Section 9 for a proof that this sequence does diverge.
3
Solutions 4.
4.1. i) an = O(bn )
ii) an = O(bn ), bn = O(an )
iii) an = O(bn )
iv) an = O(bn )
4.2. Since | sin n| ≤ 1 for all n ≥ 0, we have 2n ≤ bn ≤ 3n.
4.3. If there exist positive constants c and C such that c|an | ≤ |bn | ≤ C|an |, then 1/C |bn | ≤
|an | ≤ 1/c|bn |.
4.4. If c|an | ≤ |bn | ≤ C|an | for n ≥ N1 and c|dn | ≤ |an | ≤ |dn | for n ≥ N2 , then for
n ≥ max{N1 , N2 } we have c · e|dn | ≤ |an | ≤ C · E|dn |.
4.5. an = 1 + (−1)n , bn = 1, n ≥ 0.
4.6. If an = o(bn ) means that limn→∞ |an /bn | = 0, then an = o(bn ) and bn = o(an ) would
imply that for sufficiently large n we have both |an |/|bn | < 1 and |bn |/|an | < 1 which
lead to |an | < |bn | < |an | which is impossible.
4.7. i) By Exercise 3.3 an < 3 for all n ≥ 1, so |an /bn | < 3/n.
1
1
ii) an = n(n+1)
, so |an /bn | = n(n+1)
.
4
Solutions 5.
5.1. (i), (v), (iii), (ii), (iv). (For (i) and (v) see 5.3 below. For (v) and (iii) use the fact
that for n ≥ 1
n2
2
<
.
2
logn (n − log n)
log n
√
√
5.2. Use the fact that log2 n = log2 ( n)2 = 2 log2 n.
5.3. First show, as in Exercise 5.2, that log2 n = o(n1/j ).
5.4. For each > 0 there is an integer N such that we have both |an | < |bn | and |bn | < |cn |
and therefore |an | < 2 |cn |. Thus for < 1 we can insure |an | < 2 |cn | < |cn | (and
for ≥ 1, |an | < ( 1/2)2 |cn | < 1 · |cn | ≤ |cn |).
5.5. N = 10, N = 16.
log log 2n
log log n
log n
2
2
2
2
5.6. If an = log
, then a2n = log
= n2 . The sequence a2 , a4 , a8 , . . ., a2n , . . .
n
2n
22
has limit 0 as n → ∞. For each > 0 there is an integer N such that for all n ≥ N
2 log2 x
is decreasing for sufficiently
a2n < . Use the fact that the function f (x) = loglog
2x
n
large x (proof?) to show that if m > 2 , n ≥ N , then am < .
5.7. Adjust the argument used to show that 2n = o(n!) to show
5.8. Note that
n2n
3n
=
n
( 3/2)n
<
9
2n
for n ≥ 14.
and use result 5.6.
n
n
3n
n!
2
5.9. Use, for example, 23n2 ≤ 232n = ( 3/4)n since for n ≥ 2 we have 2n ≥ 22n which can be
proved by induction using
2
2(n+1) = 2n
2
+2n+1
2
= 2n · 22n+1 ≥ 22n · 22n+1 = 24n+1 > 22(n+1) .
5.10. For example an = n2n = (nn )2 .
5.11. The inequality holds for n = 2. Suppose it is true for some n > 1. Since then
√
1 + 1/2 + 1/3 + · · · + 1/n−1 + 1/n = (1 + 1/2 + 1/3 + · · · + 1/n−1) + 1/n ≤ 2 n + 1/n,
√
√
it is enough to show that 2 n + 1/n ≤ 2 n + 1 for n ≥ 2 to complete an argument by
induction. (Square both sides and show that √4n + n12 ≤ 4 for all n ≥ 2.)
5
Solutions 6.
6.1. (i) ∆(3n4 − 2n) = 3∆(n4 ) − 2∆n = 3[(n + 1)4 − n4 ] − 2[(n + 1) − n] = 3[(n4 + 41 n3 +
2
4
n + 41 n + 1) − n4 ] − 2[1] = 12n3 + 18n2 + 12n + 1
2
ii) ∆(7 · 2n ) = 7 · ∆(2n ) = 7[2n+1 − 2n ] = 7[2 · 2n − 2n ] = 7 · 2n
iii) ∆(n(n − 1)) = (n + 1)((n + 1) − 1) − n(n − 1) = (n + 1)n − n(n − 1) = 2n
iv) ∆(n · 2n ) = (n + 1)2n+1 − n · 2n = (n + 1) · 2 · 2n − n · 2n = (n + 2)2n
v) ∆(n!) = (n + 1)! − n! = (1 · 2 · . . . n)(n + 1) − (1 · 2 · . . . n) = n · n!
6.2. ∆(∆(3n )) = ∆(3n+1 − 3n ) = ∆((3 − 1) · 3n ) = 2∆(3n ) = 2 · 2 · 3n = 4 · 3n
6.3. ∆(g(x) + c) = ∆(g(x)) + ∆(c) = ∆(g(x)) + (c − c) = ∆(g(x))
6.4. ∆(an ) = an+1 − an = 0 implies an+1 = an .
6.5. ∆(an − bn ) = ∆(an ) − ∆(bn ) = 0, so by 6.4 an − bn = c
6.6. Not in general. For example the function
h(x) =
1 for j ≤ x < j + 1/2, j is an integer
0 for j + 1/2 ≤ x < j + 1, j is an integer
has ∆(h(x)) = 0 for all x, but is not constant. For f and g with f − g = h, ∆(f (x)) =
∆(g(x)). If, however, h(x) is a polynomial with ∆(h(x)) = 0, then h(1) = h(2) =
h(3) = . . . which implies that h(x) is a constant. So for f (x) and g(x) polynomials
∆(f (x)) = ∆(g(x)) implies f (x) − g(x) is a constant.
6.7. i) ∇(x2 ) = x2 − (x − 1)2 = 2x − 1.
ii) ∇(3n ) = 3n − 3n−1 = (3 − 1)3n−1 = 2 · 3n−1
6.8. ∆f (x) = f (x + 1) − f (x) = E(f (x)) − I(f (x)) = (E − I)(f (x))
6
Solutions 7.
7.1. i) ∆−1 (3n) = 3∆−1 = 3( 1/2(n2 + n) + c) = 3/2(n2 − n) + c0
ii) ∆−1 (n2 + 2) = ∆−1 (n2 ) + ∆−1 (2) = ( 1/3n3 − 1/2n2 + 1/6n) + 2n + c
1
· 4n + c = 1/34n + c
iii) ∆−1 (4n ) = 4−1
iv) ∆−1 (n(n − 1)) = 1/3n(n − 1)(n − 2) + c since ∆(n(n − 1)(n − 2)) = 3n(n − 1).
(See 7.6.)
7.2. i) ∆an = n, so an = 1/2n(n − 1) + c for some c. For n = 0 we have 1 = a0 =
1/2(0)(−1) + c = c, so c = 1, a = 1/2n(n − 1) + 1.
n
ii) ∆an = 3n , so an = ∆−1 (3n ) = 1/2 · 3n + c for some c. For n = 0 we have
2 = a0 = 1/2 · 30 + c = 1/2 + c, so c = 3/2. an = 1/23n + 3/2.
iii) ∆an = 3n+1 , so an = ∆−1 (3n+1 ) = 3∆−1 (3n ) = 3/2 · 3n = c. For n = 0 we have
2 = a0 = 3/2 · 30 + c = 3/2 + c, so c = 1/2. an = 3/2 · 3n + 1/2.
7.3. ∆an = n + c for some constant c, so an = 1/2n(n − 1) + cn + d for some constants c
and d. For n = 0, 2 = a0 = 1/2 · 0 + c · 0 + d = d, so d = 2. an = 1/2n(n − 1) + cn + 2
for some constant c.
7.4. i) Numbering the sequence from n = 0 we have
∆a0 = a1 − a0 = 4 − 3 = 1
∆a1 = a2 − a1 = 7 − 4 = 3
∆a2 = a3 − a2 = 12 − 7 = 5
∆a3 = a4 − a3 = 19 − 12 = 7
∆a4 = a5 − a4 = 28 − 19 = 9
Each of these is of the form ∆an = 2n + 1.
ii) If ∆an = 2n + 1, then an = ∆−1 (2n + 1) = 2 · 1/2n(n − 1) + n + c = n2 + c. With
3 = a0 = 02 + c we obtain c = 3 and an = n2 + 3. Check that this expression
does give the correct values for an , 0 ≤ n ≤ 5.
iii) If the pattern continues a6 is given by the same polynomial and a6 = 62 + 3 = 39.
7.5. For sequences ∆(an ) = 0 just for constant sequences an = c, so ∆−1 (0) = c. Therefore
∆−1 (∆−1 (0)) = ∆−1 (c) = cn + d for constants c and d.
7.6. ∆(n(n − 1)(n − 2)) = (n + 1)(n)(n − 1) − n(n − 1)(n − 2) = [(n + 1) − (n − 2)]n(n − 1) =
3n(n − 1).
7.7. For exponential sequences an = r n we have ∆an = r n+1 = r n = (r − 1)r n = (r − 1)an .
Choosing r = 4 we have ∆(4n ) = 3 · 4n .
7.8. ∆(2n ) = 2n and ∆(n2n ) = (n+2)2n (see Exercise 6.1 (iv).) Use n2n = (n+2)2n −2·2n .
7
Solutions 8.
8.1. i) ∆2 (n3 − 2n) = ∆2 (n3 ) − 2∆2 (n) = ∆(3n2 + 3n + 1) − 2∆(1) = 3(2n + 1) + 3(1) +
3(0) − 2(0) = 6n + 6.
ii) ∆2 (n(4) ) = ∆(4n(3) ) = 12n(2) = 12(n)(n − 1)
iii) ∆k n(k) = ∆k−1 (kn(k−1) ) = ∆k−2 (k(k − 1)n(k−2) = · · · = ∆(k(k − 1) · · · 2n(1) ) =
k!.
(Note: for k = 0, ∆0 = I and n(0) = 1, so ∆0 (n0 ) = 0! = 1.)
8.2. n(3) = n(n − 1)(n − 2) = n3 − 3n2 + 2n, so
n3 = n(3) + 3n2 − 2n.
Use n2 = n(2) + n(1) and n = n(1) .
8.3. As in Exercise 8.2 n(k) = n(n − 1) · · · (n − k + 1) = nk + a polynomial of degree k − 1.
So
nk = n(k) + polynomial of degree k − 1.
If each of 1, n, n2 , . . . , nk−1 can be expressed as a sum of multiples of 1, n(1) , n(2) , . . . ,
n(k−1) , then a polynomial of degree k − 1 can be obtained as such a sum.
8.4. Assuming we number from n = 0 we have a0 = 1, a1 = 7, and a2 = 19. If p(n) =
An2 + Bn + C this gives the system
a0 = A · 0 2 + B · 0 + C = 1
a1 = A · 1 2 + B · 1 + C = 7
a2 = A · 22 + B · 2 + C = 19
which has solution A = B = 3, C = 1, so
p(n) = 3n2 + 3n + 1.
8.5. i) Set an =
P
0≤j≤n
j = 0 + 1 + 2 + · · · + n, so a0 = 0. Then we have
∆an = an+1 − an = (0 + 1 + 2 + · · · + n + (n + 1)) − (0 + 1 + 2 + · · · + n) = n + 1.
1
an = ∆−1 (n+1) = ∆−1 (n)+∆−1 (1) = 1/2n(n−1)+n+c.
P Since 0 = a0 = /2(0)(−1)+
c, we have c = 0 and an = 1/2n(n + 1). Alternatively, 0≤j≤n j = An+1 − A0 , where
P
An =P∆−1 (n) = 1/2n(n − 1), so 0≤j≤n j = 1/2(n + 1)((n + 1) − 1) − 0 = 1/2n(n + 1).
1
1
1
ii)
10≤j≤n j = An+1 − A10 = /2(n)(n + 1) − /2(9)(10) = /2n(n + 1) − 45.
3
iii) By
P 8.2, n 3 = n(3) + 3n(2) + 1n(1) , so
1
0≤j≤n j = An+1 − A0 , where ∆(An ) = n(3) + 3n(2) + 1n(1) , and An = /4n(4) +
n(3) + 1/2n(2) .
8
P
j 3 = ( 1/4)(n + 1)(4) + (n + 1)(3) + 1/2(n + 1)(2) − 0 = 1/4(n4 + 2n3 + n2 )
P0≤j≤n 3
P
n
n
j
1
iv)
0≤j≤n j = An+1 − A0 , where ∆(An ) = 3 so An = /23 .
0≤j≤n 3 =
1/2(3n+1 )
− 1/2(30 ) =
3n+1 −1
.
2
8.6. ∆(4n ) = 4n+1 − 4n = (4 − 1)4n = 3 · 4n .
∆2 (4n ) = 32 · 4n
∆3 (4n ) = 33 · 4n .
8.7. i) The sequence ∆an begins with 2, 6, 10, 14, 18 which leads to the table
an
−1
1
7
17
31
49
∆an
2
6
10
14
18
∆2 a n
4
4
4
4
∆3 a n
0
0
0
an is given by a polynomial p(n) of degree 2, p(n) = An2 +Bn+C. Use p(0) = a0 = −1,
p(1) = 1, and p(2) = 7 to find p(n) = 2n2 − 1.
8.8. i) ∆(n(k) ) = ∆(n(n + 1)(n + 2) · · · (n + k − 1))
= (n + 1)(n + 2) · · · (n + k − 1)(n + k) − (n)(n + 1)(n + 2) · · · (n + k − 1)
= [(n + k) − n](n + 1)(n + 2) · · · (n + k − 1) = k(n + 1)(k−1) .
ii) ∇(n(k) ) = n(k) − (n − 1)(k) = n(n + 1) · · · (n + k − 1) − (n − 1)n(n + 1) · · · (n + k − 2)
= [(n + k − 1) − n]n(n + 1) · · · (n + k − 2) = kn(k−1) .
1
1
1
1 (k) =
8.9. ∆( 1/n(k) ) = (n+1)
(k) − /n
(n+1)(n+2)···(n+k) − n(n+1)···(n+k−1)
h
i
1
1
−k
1
= n+k
− n1 (n+1)(n+2)···(n+k−1)
= (n)(n+k)
.
(n+1)(n+2)···(n+k+1)
8.10.
1
1
−
(n + 2)(n + 3) · · · (n + k + 1) (n + 1)(n + 2) · · · (n + k)
1
1
1
−
=
n + k + 1 n + 1 (n + 2)(n + 3) · · · (n + k)
.
1
−k
=
(n + k + 1)(n + 1) (n + 2)(n + 3) · · · (n + k)
−k
= −kn(−k−1) = −kn(−(k+1))
=
(n + 1)(n + 2) · · · (n + k + 1)
∆n(−k) =
8.11. For k > 1, by Exercise 8.10,
∆−1 (n(−k) ) =
1
n(−k+1)
−k + 1
R
1
This is clearly not correct for k = 1, just as the integral rule x−k dx = −k+1
x−k+1
R
is not valid for k = 1, 1/x dx. n(−0) is not defined by the expression in Exercise 8.10.
9
1
1
1
∆−1 (n(−1) ) = ∆−1 n+1
. That is, ∆−1 n+1
= An where ∆(An ) = n+1
= An+1 −An .
1
1
If A0 = C, then A1 = A0 + 1 = c + 1, A2 = A1 + /2 = c + 1 + /2, and in general
An = c + 1 + 1/2 + · · · + 1/n. One solution, with c = 0, is the sequence of “harmonic
numbers” (see Section 9)
∆−1 (n(−1) ) = Hn = 1 + 1/2 + 1/3 + · · · + 1/n.
(What is the connection between ∆−1 (n(−1) ) and
R
x−1 dx?)
8.12. ∆2 (n(−k) ) = ∆(−kn(−k−1) = −k(−k − 1)n(−k−2) = k(k + 1)n(−(k+2)) .
8.13. Just as nj , j a non-negative integer is a sum of multiples of n(0) , n(1) , . . . , n(j) (see
Exercises 8.2, 8.3), each xj is a sum of x(0) = 1, x(1) = x, x(2) = x(x − 1), . . . , x(j) . It
follows that each polynomial f (x) of degree m can be written as
f (x) = a0 + a1 x(1) + a2 x(2) + · · · + am x(m) ,
for some choice of constants a0 , a1 , . . . , am . Suppose g(x) =
1
∆j f (0). Then
cj = j!
P
0≤j≤m cj x(j) ,
where
∆g(x) = c1 + 2c2 x(1) + 3c3 x(2) + · · · + kck x(k−1) + · · · + mcm x(m−1)
∆2 g(x) = 2c2 + 3 · 2c3 x(1) + · · · + rk(k − 1)ck x(k−2) + · · · + m(m − 1)cm x(m−2)
∆k g(x) = k!ck + (k + 1)(k) · · · 2ck+1 x(1) + · · · + m(k) cm x(m−k)
..
.
m
∆ g(x) = m!cm .
Evaluating at x = 0 gives g(0) = c0 , ∆g(0) = c1 , ∆2 g(0) = 2c2 , and in general
∆k g(0) = k!ck = k!
∆k f (0)
= ∆k f (0).
k!
Since f and g are determined uniquely by the values ∆j f (0) and ∆j g(0), 0 ≤ j ≤ m,
respectively, we have
X ∆j f (0)
f (x) = g(x) =
x(j)
j!
0≤j≤m
10
Solutions 9.
P
1 j
1 n
4
1 n
9.1. i)
0≤j≤n ( /4) = An+1 − A0 , where ∆(An ) = ( /4) . An = − /3( /4) since
∆(− 4/3( 1/4)n ) = − 4/3( 1/4 − 1)( 1/4)n = ( 1/4)n , so
X
0≤j≤n
( 1/4)j = − 4/3[( 1/4)n+1 − ( 1/4)0 ] =
( 1/4)n+1 − 1
( 1/4)n+1 − 1
=
.
1/4 − 1
− 3/4
P
P
+ 1) = 2 0≤j≤n j + 0≤j≤n 1 = 2 n(n+1)
+ n · 1 = n2 + 2n (see
2
Exercise 8.5).
P
P
P
2
n(n+1)
2n+1 −1
j
j
iii)
= 2n+1 − n2 − n2 − 1
0≤j≤n (2 − j) =
0≤j≤n 2 −
0≤j≤n j =
2−1 −
2
(see Exercise 8.5).
ii)
P
0≤j≤n (2j
R
9.2. i) Diverges. (Consider the integral x−1/2 dx
Exercise 9.7.)
P or see
j
ii) Converges. (This is a geometric series j≥0 r , with |r| < 1.)
P
iii) Converges. (Compare with j≥0 j12 .)
R
ln x
dx = ln ln x + c.)
iv) Diverges. (Consider the integral (ln x)−1 d dx
1
ln 2
v) Diverges. (Note that log n = ln n and compare with the harmonic series.
2
9.3. For example, aj =
1
j 2 +1 ,
bj =
1
j+1 .
9.4. See Exercise 9.2, (iv) and (iii).
P
9.5. Use the fact that the sequence Sn = 0≤j≤n bj is increasing
Pand has limit B to show
that Sn ≤ B for n ≥ 0. P
The sequence P
of partial sums of j≥0 aj is also increasing
and bounded by B since 0≤j≤n aj ≤ 0≤j≤n bj .
P
9.6. If the increasing sequence of partial
sums
of
j≥0 aj were bounded, the series would
P
P
be convergent. Use the fact that 0≤j≤n aj ≤ 0≤j≤n bj .
9.7.
Rn
1
1
−p
t
dt
=
−
1
for each p 6= 1. The sequence an =
p−1
1−p n
1
p > 1 and diverges for 0 < p < 1.
= 1/2 6= 0.
9.8. i) Diverges since limj→∞ j−1
2j
P
ii) Converges by comparison with j≥0
P
iii) Converges by comparison with j≥1
1
2j
1
j2
1
np−1
converges for each
since | sin j| ≤ 1.
1
since j!
≤ j12 for each j ≥ 4.
P
9.9. The series j≥1 (−1)j+1 ( 1/j ), known as the alternating harmonic series converges (see
P
P
Section 10), but j≥1 |(−1)j+1 ( 1/j )| = j≥1 1/j is the divergent harmonic series.
P
9.10. The series P
2 j≥0 |aj | converges, and since 0 ≤ aj + |aj | ≤ 2|aj | for each j ≥ 0.
The series j≥0 (aj + |aj |) therefore converges by the comparison test for series of
11
P
P
P
non-negative terms. Since j≥0 aj = j≥0 (aj + |aj |) − j≥0 |aj |, its partial sums
are the terms of a sequence which is the difference of two convergent sequences, and
hence converge.
9.11.
P
1
1≤j≤n j(j+1)
converges to 1.
=
P
1≤j≤n
1
j
−
1
j+1
=
P
1≤j≤n
∆( 1/j ) = 1 −
1
n+1 ,
so the series
9.12. ln(n!) = ln(1 · 2 · . . . · n) = ln 1 + ln 2 + · · · + ln n, where ln 1 = 0. Show that the area
beneath ln t from 1 to n is less than the sum of the areas of rectangles of width 1 and
heights ln 2, ln 3, . . . , ln n. Use integrating by parts to show that
Z
n
1
1 ln tdt = [t ln t − t]n1 = (n ln n − n) − (0 − 1).
It follows that n! > en ln n−n+1 = e ·
1
en
· (eln n )n = e( n/e)n .
9.13. limj→∞ (1 − 1/j )j = e−1 6= 0, so the series diverges.
9.14. If aj ≥ 0 for each j ≥ 0, then
P
0≤j≤n
a2j ≤
P
0≤j≤2n
aj . The terms
P
0≤j≤2n
aj
P
form a “subsequence” of the sequence of partial sums of thePconvergent series j≥0 aj ,
and hence (proof?) also converge. Therefore the series j≥0 a2j also converges by
the comparison test.
P
1
1
and a2k+1 = − k+1
for each k ≥ 0 has partial
9.15. The series j≥0 aj where a2k = k+1
P
P
2
sums
0≤j≤n aj = 0 for n odd and
0≤j≤n aj = an − n+2 for n even, and so
P
converges to 0, but j≥0 a2j is the harmonic series.
9.16. Since we have ln n < Hn < 1 + ln n, we have Hn > 1000 if n ≥ e1000 . In fact the
difference Hn − ln n has a limit as n → ∞ which is known as Euler’s constant, usually
denoted by γ, which is approximately
γ ≈ 0.5772.
It can be shown that
Hn = ln n + γ +
1
cn
1
−
+
,
2n 12n2
120n4
for some n ,
0 < n < 1.
Hn first exceeds 1000 when n is approximately e1000−γ . It can be shown that the
critical value is either be1000−γ c or de1000−γ e.
12
Solutions 10.
10.1. As in Example 22
X
0≤j≤n
Z n
X 1
X
1
1
1
1
+
<
+ 2+
dt
2
2
2
2
j +1
j
j +1 N
N t
0≤j≤N −1
N ≤j≤n
0≤j≤N −1
X
1
1
1
1
+
+
−
=
.
j2 + 1 N 2
N
n
1
≤
2
j +1
X
0≤j≤N −1
For N = 11 we have
N ≥ 10, and
P
P
1
0≤j≤10 j 2 +1
1
0≤j≤10 j 2 +1
+
as a lower bound for all partial sums
1
112
+
P
1
11
=
P
1
0≤j≤10 j 2 +1
+
12
121
P
1
0≤j≤N j 2 +1 ,
as an upper bound
for all partial sums. It follows that 0≤j≤10 j 21+1 ≈ 1.982 differs from the sum of the
series by at most 12/121 < 0.1. In fact, 2.08 is extremely close in this case.
10.2. i) Converges since limj→∞ (−1)j ln1j = 0 and ln1j is decreasing.
ii) Diverges. Note that the difference of two successive partial sums has limit 1/3:
|Sn+1 − Sn | = |an | =
n+1
,
3n + 4
while if limn→∞ Sn = L, for each positive we have |Sn+1 − Sn | = |(Sn+1 − L) −
(Sn − L)| ≤ |Sn+1 − L| + |Sn − L| < 2.
iii) Converges.
iv) Converges.
10.3. i) Let Ak = k, Bk = ∆−1 ( 1/3)k =
X
0≤k≤n
1
1 k
1/3−1 ( /3)
= − 3/2( 1/3)k . Then
X
k
=
(A
B
−
A
B
)
−
Bk+1 ∆(Ak )
n+1
n+1
0
0
3k
0≤k≤n
= [(n + 1)(− 3/2)( 1/3)n+1 − 0(− 3/2)( 1/3)0 ] −
=−
(n + 1) 1 n 1
( /3) + /2
2
=−
(n + 1) 1 n 1 1 − ( 1/3)n+1
( /3) + /2 ·
2
1 − 1/3
X
( 1/3)j
0≤j≤n
which has limit 0 + 1/2 · 3/2(1 − 0) = 3/4 =
13
1/3
.
(1− 1/3)2
X
0≤j≤n
(− 3/2)( 1/3)j+1
ii) Adding a term with j = 0 does not change the sum. Using summation by parts,
with Aj = j 2 and Bj = ∆−1 ( 1/2)j = −2( 1/2)j gives
X
X
X
j 2 ( 1/2)j = [−(n + 1)2 ( 1/2)n + 2] + 2
j( 1/2)j +
( 1/2)j
0≤j≤n
0≤j≤n
= [2 − (n + 1)2 ( 1/2)n ] +
( 1/2)n+1
1−
1 − 1/2
0≤j≤n
+2
X
j( 1/2)j .
0≤j≤n
Using summation by parts again (or results on page 68) we have
X
j 2 ( 1/2)j = [2 − (n + 1)2 ( 1/2)n ] + 2(1 − ( 1/2)n ) + [2 − (n + 2)( 1/2)n ]
0≤j≤n
whose limit is 6.
Note: One could also use j 2 = j(2) + j(1) to obtain
X
j 2 ( 1/2)j =
j≥0
X
j(2) ( 1/2)j +
j≥0
X
j( 1/2)j .
j≥0
iii) The series diverges. Its partial sums are given by
X
(An+1 Bn+1 − A1 B1 ) −
Bj+1 ∆Aj ,
1≤j≤n
where Aj = j, Bj = ∆−1 (j(2) ) = 1/3j(3) , which gives
1/3(n
X
+ 1)(n + 1)(3) − 1 · 0 − 1/3
(j + 1)(3) =
1≤j≤n
1/3(n
2
+ 1) n(n − 1) −
1/3
(j + 1)(4)
4
n+1
1
= n/12(n2 − 1)(3n + 2).
10.4. i) |x| < 1.
ii) 0 < x < 1.
iii) −1 < x < 0.
10.5. i) No. For example,
P
j √1
j
converges, but squaring the terms yields the harP
2
monic series. If, however, j≥0 aj converges and aj ≥ 0 for all j ≥ 0, then
j≥0 aj
must converge. Since limj→∞ aj = 0, there exists an integer N such that aj < 1 for
j ≥ N and therefore
for j ≥ N.
a2j < aj < 1
In this case
j≥1 (−1)
P
X
j≥0
a2j ≤
X
0≤j≤N −1
14
a2j +
X
j≥N
aj .
ii) No. Let aj = 1/j for example.
P
P
P
10.6. i) 0≤k≤n (k 2 − k) = 0≤k≤n k(2) = 0≤k≤n ∆( 1/3k(3) ) = 1/3(n + 1)(3) − 1/3 · 0
ii) Use k 2 + k = (k(2) + k(1) ) + k(1) = k(2) + 2k(1) .
10.7. i) The sum differs from 1/2(Sn + Sn−1 ) by at most 1/2|an+1 | =
1/2(S
24 + S25 ) ≈ 0.1948.
ii) Since 1/2 · 1/5! < 0.01, compute 1/2(S4 + S5 ) ≈ −0.63.
1
( 1/2)
2n+3
so set n = 24.
10.8. Both are correct.
an (bn+1 −bn )+bn+1 (an+1 −an ) = an+1 bn+1 −an bn = bn (an+1 −an )+an+1 (bn+1 −bn ).
1
1
1
10.9. i) Use (k+1)(k+2)
= k+1
− k+2
, so that the sum becomes ( 1/2 − 1/3) + ( 1/3 − 1/4) + · · · +
−1
( 1/n+1 − 1/n+2) = 1/2 − 1/n+2), or ∆(k(−1) ) = (k+1)(k+2)
which gives
X
1
=−
∆k(−1) = −[k(−1) ]n+1
1
(k + 1)(k + 2)
1≤k≤n
1≤k≤n
n+1
1
1
=−
= 1/2 −
.
k+1 1
n+2
X
ii) As in (i), use
10.10.
P
1
k(k+1)
1
1≤k≤N (k+1)(k+2)(k+3)
−2n(−3) , the sum is
X
1≤k≤N
=
1
k
=
P
−
1
k+1
or ∆(k − 1)(−1) =
1
1≤k≤N (k+3)(3)
=
P
−1
.
k(k+1)
1≤k≤N
k(−3) . Since ∆(n(−2) ) =
+1
= − 1/2[(N + 1)(−2) − 1(−2) ]
(− 1/2)∆k(−2) = − 1/2[k(−2) ]N
1
= − 1/2
1
1
−
(N + 2)(N + 3) 2 · 3
The limit as N → ∞ is (− 1/2)(− 1/6) = 1/12.
10.11. i) 23/24, 13/120, 93/3240, 37/6720
2
+11k+6
6
Note that bk = k3 (k6k
3 +6k 2 +11k+6) < k 4 , so the sequence (bk ) tends to 0 more rapidly
P
1
1 3
than k13 . (The sequence (k−1)k(k+1)
− k13 does so even more rapidly.)
k≥1 /k =
P
P
P
1
k≥1 k(−3) +
k≥1 bk = /12 +
k≥1 bk by Exercise 10.10. Obtain the estimate by
adding the first few terms of the sequence (bk ).
15
10.12.
1
k3
=
1
(k−1)(k)(k+1)
=
−1
k3 (k2 −1)
which leads to (for k > 1)
X 1
X
X 1
X
1
1
−
=
1
+
=
1
+
3
3
3
2
k
k
(k + 1)(3)
k (k − 1)
k≥1
k≥2
k≥2
k≥2
= 1 + 1/4 −
≈
5/4
−
10.13. Let Aj = Hj and ∆Bj = j. Then ∆Aj =
by parts yields
X
1≤j≤n
jHj =
1/2 j
(2) Hj
k≥2
1/24
−
1
k 3 (k 2
.
− 1)
− 1/216 − 1/960 ≈ 1.203
1
j+1
n+1
1
X
= j(−1) and Bj = 1/2 j(2) . Summation
X
1/2 j
(−1) (j
1≤j≤n
= [ 1/2(n + 1)nHn − 0] − 1/2
X
1≤j≤n
= 1/2(n + 1)nHn − 1/4 n(n + 1)
= 1/4 n(n + 1)[2Hn − 1].
16
j
+ 1)(2)